Chapter 2: Introduction to Systems Problem 2.1 (i)

The currents flowing out of node 1 along resistors R1, R2, and capacitor C, are given by

iR1 =

y (t ) − v (t ) , R1

iR 2 =

y (t ) , R2

iC = C

dy dt

Applying the Kirchoff’s current law to node 1 and summing up all the currents, gives y (t ) − v(t ) y (t ) dy + +C =0 R1 R2 dt

or,

C

dy  1 1  v(t ) +  +  y (t ) = dt  R1 R2  R1

dy R1 + R2 1 + y (t ) = v(t ) . dt CR1 R2 CR1

or,

(ii) (a) Linear: For v1(t) applied as the input, the output y1(t) is given by dy1 R1 + R2 dy R + R2 1 y1 (t ) = v1 (t ) ⇒ v1 (t ) = CR1 1 + 1 + y1 (t ) dt CR1 R2 CR1 dt R2 For v2(t) applied as the input, the output y2(t) is given by dy 2 R1 + R2 dy R + R2 1 + y 2 (t ) = v 2 (t ) ⇒ v 2 (t ) = CR1 2 + 1 y 2 (t ) . dt CR1 R2 CR1 dt R2

For v3(t) = α v1(t) + β v2(t) applied as the input, the output y3(t) is given by dy 3 R1 + R2 1 (αv1 (t ) + βv 2 (t ) ) . + y 3 (t ) = dt CR1 R2 CR1 Substituting the values of v1(t) and v2(t) from the earlier equations, we get

 dy   dy  dy 3 R1 + R2 R + R2 R + R2 + y 3 (t ) = α  1 + 1 y1 (t ) + β  2 + 1 y 2 (t )  . dt CR1 R2 CR1 R2 CR1 R2  dt   dt  Rearranging the on the right hand side of the equation, we get dy 3 R1 + R2 d (αy1 + β y 2 ) R1 + R2 (αy1 (t ) + βy 2 (t ) ) , + y 3 (t ) = + dt CR1 R2 dt CR1 R2

which implies that

y 3 (t ) = αy1 (t ) + βy 2 (t ) .

The system is, therefore, linear. (b) Time-invariance: For v(t – t0) applied as the input, the output y1(t) is given by

(S2.1)

42

Chapter 2 dy1 R1 + R2 1 + y1 (t ) = v(t − t 0 ) . dt CR1 R2 CR1

Substituting τ = t – t0 (which implies that dt = dτ), we get dy1 (τ + t 0 ) R1 + R2 1 + y1 (τ + t 0 ) = v(τ) . dτ CR1 R2 CR1 Comparing with Eq. (S2.1), we get

y (τ) = y1 (τ + t0 ) or,

y1 (τ) = y ( τ − t0 ) ,

proving that the system is time-invariant. (c) Memoryless: Express Eq. (S2.1) as y (t ) =

1 CR1

t

∫

v(τ)dτ −

−∞

R1 + R2 CR1 R2

t

∫ y(τ)dτ .

−∞

The output y(t) at t = t0 is given by

y (t ) t =t = 0

1 CR1

t0

∫

v(τ)dτ −

−∞

R1 + R2 CR1 R2

t0

∫ y(τ)dτ

−∞

From the first integral on the right hand side of the equation, it is clear that all previous values of the input v(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. The system has, therefore, a memory and is not memoryless. (d) Causal: From the previous result, we deduce that the system is causal since only the past values of the input v(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. (e) Invertible: The system is invertible as v(t) can be determined from the following relationship v(t ) = CR1

dy R1 + R2 + y (t ) . dt R2

(f) Stable: The system is BIBO stable since a bounded input will always produce a bounded output as shown below. Using Theorem 3.1, the output of the system defined by Eq. (S2.1) is given by y (t ) = e − p

t

∫e

p

−∞

[

p (t ) = exp −

where

1 v ( τ) dτ × CR 1

R1 + R2 CR1 R2

]

t .

From the above solution, it is clear that the output y(t) is bounded in the input v(t) is bounded. Problem 2.2

(a)

The currents flowing in resistor R, inductor L, and capacitor C, are given by t

iR =

y (t ) 1 dy , iL = ∫ y (τ ) dτ , iC = C . R L −∞ dt

▌

Solutions

43

Applying the Kirchoff’s current law, we obtain t

y (t ) 1 dy + ∫ y (τ )dτ + C = i (t ) , R L −∞ dt

C

or, by differentiating,

d 2 y 1 dy 1 di + + y (t ) = 2 dt R dt L dt

d2y 1 dy 1 1 di + + y (t ) = . 2 dt RC dt LC C dt

or,

(S2.2)

(ii) (a) Linear: For i1(t) applied as the input, the output y1(t) is given by

d 2 y1 dt 2

+

1 dy1 1 1 di1 + y1 (t ) = RC dt LC C dt

⇒

di1 d 2 y1 1 dy1 1 =C + + y1 (t ) dt R dt L dt 2

For i2(t) applied as the input, the output y2(t) is given by d 2 y2 dt 2

+

1 dy 2 1 1 di 2 + y 2 (t ) = RC dt LC C dt

⇒

di 2 d 2 y 2 1 dy 2 1 =C + + y 2 (t ) . dt R dt L dt 2

For i3(t) = α i1(t) + β i2(t) applied as the input, the output y3(t) is given by d 2 y3 dt or,

d 2 y3 dt

2

+

2

+

1 dy 3 1 1 d (αi1 + βi 2 ) + , y 3 (t ) = RC dt LC C dt

1 dy 3 1 α di1 β di 2 + + y 3 (t ) = RC dt LC C dt C dt

Substituting the values of di1/dt and di2/dt from the earlier equations, we get d 2 y3 dt 2

+

d 2 y  d 2 y  1 dy 3 1 1 dy1 1 1 dy 2 1 + + + y 3 (t ) = α  21 + y1 (t ) + β  2 2 + y 2 (t ) . RC dt LC RC dt LC RC dt LC  dt   dt 

Rearranging the on the right hand side of the equation, we get d 2 y3 dt 2

+

d 2 (αy1 + β y 2 ) 1 dy 3 1 1 d (αy1 + β y 2 ) 1 (αy1 + βy 2 ) . + y 3 (t ) = + + 2 RC dt LC RC dt LC dt y 3 (t ) = αy1 (t ) + β y 2 (t ) .

Comparing with (S2.2) implies that The system is therefore linear.

(b) Time-invariance: For i1(t – t0) applied as the input, the output y1(t) is given by d 2 y1 dt

2

+

1 dy1 1 1 di1 (t − t 0 ) y1 (t ) = + . RC dt LC C dt

Substituting τ = t – t0 (which implies that dt = dτ), we obtain d 2 y1 (τ + t 0 ) dτ

2

+

1 dy1 (τ + t 0 ) 1 1 di1 ( τ) + y1 (τ + t 0 ) = . RC dτ LC C dτ

44

Chapter 2

Comparing with Eq. (S2.2.2), we obtain y (τ) = y1 (τ + t0 ) or,

y1 (τ) = y ( τ − t0 ) ,

proving that the system is time-invariant. (c) Memoryless: Express Eq. (S2.2) as y (t ) =

1 C

t

∫

i (α )dα −

−∞

1 LC

τ

t

∫ ∫

y (α )dαdτ −

−∞ −∞

1 RC

t

∫ y(α)dα .

−∞

The output y(t) at t = t0 is given by y (t ) t =t = 0

1 C

t0

∫

i ( α ) dα −

−∞

1 LC

t0

τ

∫ ∫

y (α )dαdτ −

−∞ −∞

1 RC

t0

∫ y ( α ) dα

−∞

From the first integral on the right hand side of the equation, it is clear that all previous values of the input i(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. The system has, therefore, memory and is not memoryless. (d) Causal: From the previous result, we deduce that the system is causal since only the past values of the input i(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. (e) Invertible: The system is invertible as i(t) can be determined from the following relationship d 2 y 1 dy 1 di =C 2 + + y (t ) . dt R dt L dt (f) Stable: The system is BIBO stable since a bounded input will always produce a bounded output.

▌

Problem 2.3

(i)

For input x1(t), x2(t), and αx1(t) + βx2(t), the respective outputs are given by

x1 ( t ) → c1 x1 (t ) + c2 x12 (t ) = y1 ( t ) x2 ( t ) → c1 x2 (t ) + c2 x22 (t ) = y2 ( t ) α x1 ( t ) + β x2 ( t ) → c1 α x1 ( t ) + β x2 ( t )  + c2 α x1 ( t ) + β x2 ( t )  = y(t ) 2

Because y(t ) ≠ α y1 (t ) + β y2 (t ) , the demodulator is a non-linear device. (ii) (a) Time Invariance: For input x1(t) and x2(t) = x1(t − T), the respective outputs are given by

x1 ( t ) → c1 x1 (t ) + c2 x12 (t ) = y1 ( t ) x2 ( t ) = x1 ( t − T ) → c1 x2 (t ) + c2 x22 (t ) = c1 x1 ( t − T ) + c2 x12 ( t − T ) = y2 ( t ) Since y1 ( t − T ) = y2 ( t ) , the system is time invariant. (b) Memory: The output v2(t) depends only on the current value of the input v1(t). Therefore, the system is memoryless. All memoryless systems are causal. Therefore, the system is also causal. (c) Invertible: The input to the system, v1(t) can be calculated using the following equation

Solutions

45

−c1 ± c12 + 4c2 v2 (t ) . v1 (t ) = 2c2

For a given value of v2(t), two possible values of v1(t) exist As the input v1(t) cannot be uniquely determined from the output v2(t), the system is NOT invertible. (d) Stable: Assuming that |v1(t)| ≤ M < ∞, the output v2(t) is bounded by v2 ( t ) = c1v1 (t ) + c2 v12 (t ) < c1 v1 (t ) + c2 v12 (t ) < c1M + c2 M 2 < ∞ .

Therefore, the system is stable.

▌

Problem 2.4

(i)

The modulated signal is given by

s ( t ) = A [1 + km(t )] cos ( 2π f c t ) = 5 [1 + 2ksin(200π t ) ] cos ( 2000000π t ) .

1 + 2ksin(200π t ) ≥ 0 implies (ii)

2ksin(200π t ) ≥ −1 , or, k ≤ 0.5 .

Any value of k in the range (0 ≤ k ≤ 0.5) can be used. We use k = 0.5 in the rest of the problem. The AM signal is given by

s (t ) = A [1 + km(t ) ] cos ( 2π f c t ) = 5 [1 + 0.8sin(200π t ) ] cos ( 2000000π t ) . (iii) Expanding the above equation, we get

s (t ) = 5cos ( 2000000π t ) + 4sin(200π t ) cos ( 2000000π t ) = 5cos ( 2000000π t ) + 2sin(2000200π t ) + 2sin(1999800π t ) It is observed that the AM signal has three frequency components at 1,000,000 Hz, 1,000,200 Hz, and 999,900 Hz. The frequency component at 1,000,000 Hz represents the carrier signal, while the remaining two frequency components at 1,000,200 Hz and 999,900 Hz represent the sinusoidal tone. Therefore, the frequency of the sinusoidal tone is shifted to a higher frequency range. ▌ Problem 2.5

(i)

Dividing both sides by M, Eq. (2.16) can be expressed as d2y 1 r dy k y (t ) = x(t ) . + + M dt M M dt

Comparing with the given expression, the coefficients are given by k M ωn r = Q M ω n2 =

and (ii)

k M kM Q= . r

⇒ ωn = ⇒

k , the natural frequency ωn can be increased either by: (a) increasing the value of M the spring constant k, or (b) by decreasing the value of the mass M.

Since ωn =

46

Chapter 2

kM , the value of Q can be reduced either by: (a) reducing the value of the spring r constant k, (b) reducing the value of mass M, or (c) by increasing the value of r.

Since Q =

(iii)

d2y dy +δ + ε y (t ) = γ x(t ) , 2 dt dt

with δ =

r k 1 ,ε = ,γ = M M M

(S2.5.1)

(a) Linear: For x1(t) applied as the input, the output y1(t) is given by

d 2 y1 dy + δ 1 + ε y1 (t ) = γ x1 (t ) ⇒ 2 dt dt

x1 (t ) = γ1

d 2 y1 δ dy1 ε + + y1 (t ) dt 2 γ dt γ

x2 (t ) = γ1

d 2 y2 δ dy2 ε + + y2 (t ) . dt 2 γ dt γ

For x2(t) applied as the input, the output y2(t) is given by

d 2 y2 dy + δ 2 + ε y2 (t ) = γ x2 (t ) ⇒ 2 dt dt

For x3(t) = αx1(t) + βx2(t) applied as the input, the output y3(t) is given by

d 2 y3 dy + δ 3 + ε y3 (t ) = γ x3 (t ) = γ (α x1 (t ) + β x2 (t ) ) . 2 dt dt Substituting the values of x1 (t ), x2 (t ) from the earlier equations, we obtain

 d 2 y1   d 2 y2  d 2 y3 dy3 dy1 dy δ ε y ( t ) α δ ε y ( t ) β + + = + + + + δ 2 + ε y2 (t )  3 1    2 2 2 dt dt dt dt  dt   dt  =

d 2 (α y1 + β y2 ) d (α y1 + β y2 ) +δ + ε (α y1 (t ) + β y2 (t ) ) 2 dt dt

Comparing the left-hand and right-hand sides, we obtain

y 3 (t ) = αy1 (t ) + βy 2 (t ) . The system is therefore linear. (b) Time-invariance: For x1(t) applied as the input, the output y1(t) is given by

d 2 y1 dy + δ 1 + ε y1 (t ) = γ x1 (t ) 2 dt dt

(S2.5.2)

For x2(t) =x1(t – t0) applied as the input, the output y2(t) is given by

d 2 y2 dy + δ 2 + ε y2 (t ) = γ x2 (t ) = γ x1 (t − t0 ) . 2 dt dt Substituting τ = t + t0 (which implies that dt = dτ) in Eq. (S2.5.2), we obtain

d 2 y1 (τ − t0 ) dy (τ − t0 ) +δ 1 + ε y1 (τ − t0 ) = γ x1 (τ − t0 ) 2 dτ dτ Substituting t = τ , we obtain

d 2 y1 (t − t0 ) dy (t − t0 ) +δ 1 + ε y1 (t − t0 ) = γ x1 (t − t0 ) 2 dt dt

(S2.5.3)

Solutions

47

Comparing with Eq. (S2.5.2), we obtain

y2 (t ) = y1 (t − t0 ) , proving that the system is time-invariant. (c) Memoryless: Express Eq. (S2.5.1) as

y (t ) = γ

τ

t

∫∫

x(α )dα dτ − ε

−∞ −∞

τ

t

∫∫

x(α )dα dτ − δ

−∞ −∞

t

∫ y(α )dα .

−∞

The output y(t) at t = t0 is given by

y (t ) t =t = γ 0

t0 τ

∫∫

x(α )dα dτ − ε

−∞ −∞

t0 τ

∫∫

x(α ) dα dτ − δ

−∞ −∞

t0

∫ y(α )dα

−∞

From the first integral on the right hand side of the equation, it is clear that all previous values of the input x(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. The system has, therefore, memory and is not memoryless. (d) Causal: From the previous result, we deduce that the system is causal since only the past values of the input x(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. (e) Invertible: The system is invertible as i(t) can be determined from the following relationship

x(t ) = γ1

d 2 y δ dy ε + + y (t ) . dt 2 γ dt γ

(f) Stable: The system is BIBO stable since a bounded input will always produce a bounded output.

▌

Problem 2.6

(i)

Substituting, t = k∆t in

d2y dy +5 + 6 y (t ) = 0 , yields dt dt

d2y dt

+5 t = k∆t

dy dt

t = k∆t

+ 6 y (t ) t = k∆t = 0 .

Substituting the values of the first and second derivative from the backward finite difference scheme, we get

y (k ∆t ) − 2 y ((k − 1)∆t ) + y ((k − 2)∆t )

( ∆t )

2

+5

y (k ∆t ) − y ((k − 1)∆t ) + 6 y (k ∆t ) = 0 , ∆t

or, y ( k ∆t ) − 2 y (( k − 1) ∆t ) + y (( k − 2) ∆t ) + 5 [ y ( k ∆t ) − y (( k − 1)∆t ) ] ∆t + 6 y ( k ∆t )(∆t ) 2 = 0 . Substituting y(k∆t) = y[k], y((k – 1)∆t) = y[k − 1], and y((k – 2)∆t) = y[k − 2], the above equation reduces to

y[k ] − 2 y[k − 1] + y[k − 2] + 5 { y[k ] − y[ k − 1]} ∆t + 6 y[k ](∆t ) 2 = 0 , which simplifies to

48

Chapter 2

(1 + 5∆t + 6(∆t ) )y[k ] + (− 2 − 5∆t )y[k − 1] + y[k − 2] = 0 . 2

(ii)

Substituting the value of the first CT initial condition y(0) = 3 in

y[k] = y(k∆t) for k = 0, we obtain

y[0] = y(0) = 3.

Similarly, substituting the value of the second CT initial condition y (0) = −7 in

dy dt

≈ t = k∆t

y (0) =

for k = 0, we get

y (k∆t ) − y ((k − 1)∆t ) ∆t

y[0] − y[−1] = −7 . ∆t

Simplifying the above, we obtain

y[−1] = y[0] − ∆ty (0) = 3 + 7∆t . (iii) By substituting ∆t = 0.02s in the difference equation,

(1 + 5∆t + 6(∆t ) )y[k ] + (− 2 − 5∆t )y[k − 1] + y[k − 2] = 0 , 2

(1.1024) y[k ] + (− 2.1)y[k − 1] + y[k − 2] = 0 , y[k ] = 1.90493 y[k − 1] − 0.90711y[k − 2]

we obtain or,

with y[0] = 3 and y[−1] = 3 + 7(0.02) = 3.14. The Matlab code used to compute and plot the DT solution is given in Program 2.6. For comparison, we also plot the CT solution. The two plots are included in Fig. S2.6.

CT Solution of Problem 2.6 3

yct

2

1 0

0

0.1

0.2

0.3

0

5

10

15

0.4

0.5 0.6 0.7 t DT Solution of Problem 2.6

0.8

0.9

40

45

3

ydt

2 1 0

20 25 30 k: Recall t = k ×∆ t

35

Fig. S2.6: Plots obtained from the CT (top) and DT (bottom) solution of the 2nd order differential equation in Problem 2.6.

Solutions

49

Program 2.6: MATLAB code for Problem 2.6. % MATLAB code for Problem 2.6 % plot CT result N = 50; % No of points to be plotted k = 0:N-1; % time index dt = 0.02; % discretization step t = k*dt; % time instants yct = exp(-3*t)+2*exp(-2*t); % CT answer subplot(2,1,1) plot(t,yct), grid on xlabel('t') % Label of X-axis ylabel('yct') % Label of Y-axis title('CT Solution of Problem 2.6'); axis ([0 max(t) 0 max(yct)]) % compute finite difference approximation ydt(1) = 3+7*dt; % this is actually y[-1] ydt(2)= 3; % this is actually y[0] for i = 3:N+1 ydt(i) = 1/(1+5*dt+6*dt^2)*((2+5*dt)*ydt(i-1)- ydt(i-2)); end subplot(2,1,2) stem(k,ydt(2:N+1),'fill'), grid on xlabel('k: Recall t = k\times\Deltat');% Label of X-axis ylabel('ydt') % Label of Y-axis title('DT Solution of Problem 2.6'); axis ([0 max(k) 0 max(ydt)]) Problem 2.7

Starting from the initial value x(0) = 0V, The output of the delta modulator receiver is computed from the recursive expression xˆ (kT ) = xˆ ((k − 1)T ) + bk ∆ where bk is 1 if bit 1 is received, while bk is −1 if bit 0 is received. The computed values are shown in Table S2.7 with the waveform plotted in Fig. S2.7. ▌ Table S2.7: Decoded output of the delta modulator in Problem 2.7 k

0

t

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.0

bk xˆ (t )

1 1

2 1

3 1

4 1

5 1

6 0

7 1

8 1

9 1

10 11 12 13 14 15 16 17 18 19 20 1

1

1

0

0

0

0

0

0

0

0

0 0.1 0.2 0.3 0.4 0.5 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2

50

Chapter 2

Reconstructed waveform for x(t) 1 0.9 0.8 0.7

x(t)

0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.2

0.4

0.6

0.8

1 time (t)

1.2

1.4

1.6

1.8

2

Fig. S2.7:Waveform reconstructed from the delta-modulated bit stream in Problem 2.7. Problem 2.8

Eq. (2.27) represents a low filter that averages the four neighboring pixels to compute the value of the reference pixel located at (m, n). Specifically, y[m, n] =

1 (x[m, n] + x[m, n − 1] + x[m − 1, n] + x[m − 1, n − 1]) . 4

The system is invertible as the input x[m, n] can be constructed in a recursive way from the current value of the input and the previously reconstructed values of the input pixels based on the following relationship

x[m, n] = 4 y[m, n] − x[m, n − 1] − x[m − 1, n] + x[m − 1, n − 1] . Since both the original filter and the inverse system use the values of the past pixels to compute the reference pixel, both systems are causal in nature. ▌ Problem 2.9

(i)

y (t ) = x (t − 2)

(a) Linearity: Since

x1 (t ) → x1 (t − 2) = y1 (t ) x2 (t ) → x2 (t − 2) = y2 (t ) α x1 (t ) + β x2 (t ) → α x1 (t − 2) + β x2 (t − 2) = α y1 (t ) + β y2 (t ) therefore, the system is a linear system. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

x1 (t ) → x1 (t − 2) = y1 (t )

x2 (t ) = x1 (t − T ) → x2 (t − 2) = x1 (t − T − 2) = y2 ( t )

and y1 (t − T ) = x1 (t − T − 2) = y2 (t ) , the system is time invariant.

Solutions

51

(c) Stability: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = x(t − 2) ≤ M is also bounded proving that the system is BIBO stable. (d) Causality: Since the output depends only on the past input and does not depend on the future values of the input, therefore, the system is causal. (ii)

y (t ) = x(2t − 5) (a) Linearity: Since

x1 (t ) → x1 (2t − 5) = y1 ( t ) x2 (t ) → x2 (2t − 5) = y2 ( t ) α x1 (t ) + β x2 (t ) → α x1 (2t − 5) + β x2 (2t − 5) = α y1 (t ) + β y2 (t ) therefore, the system is a linear system. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

x1 (t ) → x1 (2t − 5) = y1 (t )

x2 (t ) = x1 (t − T ) → x2 (2t − 5) = y2 ( t )

which implies that

x2 (t ) = x1 (t − T ) → y2 (t ) = x2 (2t − 5) x

2

( t ) = x1 (t −T )

= x1 ( 2t − 5 − T ) .

On the other hand,

y1 (t − T ) = x1 (2t − 5) t =t −T = x1 (2(t − T ) − 5) = x1 (2t − 2T − 5) , and y1 (t − T ) ≠ y2 ( t ) . Therefore, the system is NOT time invariant. (c) Stability: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = x( 2t − 5) ≤ M is also bounded proving that the system is BIBO stable. (d) Causality: For (t > 5), the output depends on the future values of the input, therefore, the system is NOT causal. (iii)

y (t ) = x(2t ) − 5 (a) Linearity: Since

x1 (t ) → x1 (2t ) − 5 = y1 ( t ) x2 (t ) → x2 (2t ) − 5 = y2 ( t ) α x1 (t ) + β x2 (t ) → α x1 (2t ) + β x2 (2t ) − 5 ≠ α y1 (t ) + β y2 (t ) because α y1 (t ) + β y2 (t ) = α x1 (2t ) + β x2 (2t ) − 5(α + β ) . Therefore, the system is NOT linear. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

52

Chapter 2 x1 (t ) → x1 (2t ) − 5 = y1 (t )

x2 (t ) = x1 (t − T ) → x2 (2t ) − 5 = y2 ( t )

which implies that

x2 (t ) = x1 (t − T ) → y2 (t ) = x2 (2t ) − 5 x

2

(t ) = x1 (t −T )

= x1 (2t − T ) − 5 .

On the other hand,

y1 (t − T ) = x1 (2t ) t =t −T − 5 = x1 (2(t − T )) − 5 = x1 (2t − 2T ) − 5 . Clearly, y1 (t − T ) ≠ y2 (t ) , therefore, the system is NOT time invariant. (c) Stability: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = x(2t ) − 5 ≤ x(2t ) + 5 ≤ M + 5 is also bounded proving that the system is BIBO stable. (d) Causality: For (t > 0), the system requires future values of the input to calculate the current value of the input. Therefore, the system is NOT causal. (iv)

y (t ) = tx (t + 10)

(a) Linearity: Since

x1 ( t ) → tx1 ( t + 10 ) = y1 ( t ) x2 ( t ) → tx2 ( t + 10 ) = y2 ( t ) α x1 ( t ) + β x2 ( t ) → α tx1 ( t + 10 ) + β tx2 ( t + 10 ) = α y1 (t ) + β y2 (t ) therefore, the system is a linear system. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

x1 (t ) → tx1 (t + 10) = y1 (t ) x2 (t ) = x1 (t − T ) → tx2 (t + 10) = tx1 (t − T + 10) = y2 (t ) We also note that

y1 ( t − T ) = (t − T ) x1 ( t − T + 10 ) ≠ y2 ( t ) ,

therefore, the system is NOT time invariant. (c) Stability: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = tx(t + 10) = t x(t + 10) ≤ M t is unbounded as t → ∞. Therefore, the system is NOT BIBO stable. (d) Causality: Since the output depends on the future values of the input, and therefore the system is NOT causal. (v)

 2 y (t ) = 2u ( x ( t ) ) =   0

(a) Linearity: Since

x (t ) ≥ 0 x (t ) < 0

Solutions

53

x1 (t ) → 2u ( x1 ( t ) ) = y1 (t ) x2 (t ) → 2u ( x2 ( t ) ) = y2 (t ) α x1 (t ) + β x2 (t ) → 2u (α x1 ( t ) + β x2 ( t ) ) = y (t )

(

)

(

)

and α y1 (t ) + β y2 (t ) = 2α u x1 ( t ) + 2 β u x2 ( t ) ≠ y(t ) . Therefore, the system is NOT linear. Also, we note that

y2 (t ) − y1 (t ) = 2u ( x2 ( t ) ) − 2u ( x1 ( t ) ) ≠ λ [ x2 (t ) − x1 (t )] . Therefore, the system is NOT an incrementally linear system either. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

x1 (t ) → 2u ( x1 ( t ) ) = y1 (t ) x2 (t ) = x1 (t − T ) → 2u ( x2 ( t ) ) = 2u ( x1 ( t − T ) ) = y2 (t ) We note that y1 (t − T ) = 2u ( x1 ( t − T ) ) = y2 (t ) , therefore, the system is time invariant. (c) Stability: Since |y(t)| ≤ 2, therefore, the system is BIBO stable. (d) Causality: The output at any time instant does not depend on future value of the input. The system is, therefore, causal. (vi)

 0 y (t ) =   x(t ) − x(t − 5)

t<0 t≥0

= [ x(t ) − x(t − 5) ] u (t )

(a) Linearity: Since

x1 (t ) → [ x1 (t ) − x1 (t − 5)] u (t ) = y1 (t ) x2 (t ) → [ x2 (t ) − x2 (t − 5)] u (t ) = y2 (t ) α x1 (t ) + β x2 (t ) → {α x1 (t ) + β x2 (t )} − {α x1 (t − 5) + β x2 (t − 5)} u (t ) = y(t )

=α [ x1 (t ) − x1 (t − 5)] u (t ) + β [ x1 (t ) − x1 (t − 5)] u (t ) = α y1 (t ) + β y2 (t ),

the system is linear. (b) Time Invariance: For inputs x1(t) and x2(t) = x1(t − T), the outputs are given by

x1 (t ) → [ x1 (t ) − x1 (t − 5)] u (t ) = y1 (t ) x2 (t ) = x1 (t − T ) → [ x2 (t ) − x2 (t − 5)] u (t ) = y2 (t ) We note that y1 (t − T ) ≠ y2 ( t ) since

y1 (t − T ) = y1 (t ) t =t −T = [ x1 (t ) − x1 (t − 5)] u (t ) t =t −T = [ x1 (t − T ) − x1 (t − T − 5)] u (t − T ) and

y2 (t ) = [ x2 (t ) − x2 (t − 5)] u (t ) == [ x1 (t − T ) − x1 (t − T − 5)] u (t ) .

Therefore, the system is NOT time invariant.

54

Chapter 2 (c) Stablity: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = x(t ) − x(t − 5) ≤ x(t ) + x(t − 5) ≤ 2 M is also bounded. Therefore, the system is BIBO stable. (d) Causality: The output does not depend on the future values of the input, therefore, the system is causal. (vii) y (t ) = 7 x 2 (t ) + 5 x(t ) + 3 (a) Linearity: For x1(t) applied as the input, the output y1(t) is given by

y1 (t ) = 7 x12 (t ) + 5 x1 (t ) + 3 For x2(t) applied as the input, the output y2(t) is given by

y2 (t ) = 7 x22 (t ) + 5 x2 (t ) + 3 . For x3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by

or,

(

y3 (t ) = 7(αx1 (t ) + βx2 (t ) )2 + 5(αx1 (t ) + βx2 (t ) ) + 3 ,

) (

)

y3 (t ) = α 7 x12 (t ) + 5 x1 (t ) + 3 + β 7 x12 (t ) + 5 x2 (t ) + 3 + 14αβ x1 (t ) x2 (t ) + 3(1 − α − β) y1 ( t )

y 2 (t )

The above result implies that

y3 (t ) ≠ αy1 (t ) + βy2 (t ) , And hence the system is not linear. (b) Time Invariance: For x1(t) and x2(t) applied as the inputs, the outputs are given by

x1 (t ) → 7 x12 (t ) + 5 x1 (t ) + 3 = y1 (t ) x2 (t ) = x1 (t − T ) → 7 x22 (t ) + 5 x2 (t ) + 3 = y2 (t ). Substituting x2 (t ) = x1 (t − T ) we obtain,

y2 (t ) = 7 x12 (t − T ) + 5 x2 (t − T ) + 3 . We also note that

y1 (t − T ) = 7 x12 (t − T ) + 5 x2 (t − T ) + 3 ,

implying that y1 (t − T ) = y2 (t ) . The system is, therefore, time invariant. (c) Stablity: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) = 7 x 2 (t ) + 5 x(t ) + 3 ≤ 7 x(t ) x(t ) + 5 x(t ) + 3 ≤ 7 M 2 + 5M + 3 is also bounded. Therefore, the system is BIBO stable. (d) Causality: The output y(t) at t = t0 requires only one value of the input y(t) at (t = t0). Therefore, the system is causal. (viii) y(t) = sgn(x(t)) (a) Linearity: For x1(t) applied as the input, the output y1(t) = sgn(x1(t)). For x2(t) applied as the input, the output y2(t) = sgn(x2(t)).

Solutions

55

For x3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by

y3 (t ) = sgn (αx1 (t ) + βx2 (t ) ) ≠ α sgn (x1 (t ) ) + β sgn (x2 (t ) ) . The above result implies that

y3 (t ) ≠ αy1 (t ) + βy2 (t ) , And hence the system is not linear. (b) Time Invariance: For x1(t) and x2 (t ) = x1 (t − T ) applied as the inputs, the outputs are given by

x1 (t ) → sgn (x1 (t ) ) = y1 (t ) x2 (t ) = x1 (t − T ) → sgn (x2 (t ) ) = y2 (t ) Substituting x2 (t ) = x1 (t − T ) we obtain,

y2 (t ) = sgn (x1 (t − T ) ) . y1 (t − T ) = sgn (x1 (t − T ) ) ,

We also note that

implying that y1 (t − T ) = y2 (t ) . The system is, therefore, time invariant. (c) Stablity: The system is stable as the output is always bounded between the values of −1 and 1. (d) Causality: The output y(t) at (t = t0) requires only one value of the input x(t) at (t = t0), therefore, the system is causal. t0

(ix)

y (t ) =

∫ x(λ)dλ + 2 x(t )

−t0

(a) Linearity: For x1(t) and x2(t) applied as the inputs, the outputs are given by t0

∫ x1 (λ)dλ + 2 x1 (t ) ,

y1 (t ) =

−t0

t0

y2 (t ) =

∫ x2 (λ)dλ + 2 x2 (t ) .

−t0

For x3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by

y3 (t ) =

t0

∫ (α x (λ ) + β x (λ ) ) d λ + 2 (α x (t ) + β x (t ) ) 1

2

1

2

− t0

 t0   t0  = α  ∫ x1 (λ )d λ + 2 x1 (t )  + β  ∫ x2 (λ )d λ + 2 x2 (t )  ,  −t0   − t0  y1 ( t )

y2 ( t )

= α y1 (t ) + β y2 (t ) Therefore, the system is linear. (b) Time Invariance: For x1(t) and x2 (t ) = x1 (t − T ) applied as the inputs, the outputs are given by

56

Chapter 2

t0

x1 (t ) → y1 (t ) =

∫ x1 (λ)dλ + 2 x1 (t )

−t0

t0

x2 (t ) = x1 (t − T ) → y2 (t ) =

∫ x2 (λ)dλ + 2 x2 (t ) ,

−t0

Substituting x2 (t ) = x1 (t − T ) we obtain, t0

y2 (t ) =

∫ x (λ − T )d λ + 2 x (t − T ) . 1

1

− t0

t0 −T

By substituting λ´ = λ – T, we get y2 (t ) =

∫

x1 (λ ′)d λ ′ + 2 x1 (t − T ) .

− t0 −T

t0

y1 (t − T ) =

We also note that

∫ x1 (λ)dλ + 2 x1 (t − T ) ,

−t0

implying that y1 (t − T ) ≠ y2 (t ) . The system is, therefore, NOT time invariant. (c) Stablity: Assume that the input is bounded |x(t)| ≤ M. Then, the output

y (t ) =

t0

t0

−t0

−t0

∫ x(λ)dλ + 2 x(t ) ≤ ∫ x(λ) dλ + 2 x(t ) ≤ 2Mt0 + 2M

is also bounded. Therefore, the system is BIBO stable. (d) Causality: To solve the integral, the output y(t) always requires the values of the input x(t) within the range (−t0 ≤ t ≤ t0) no matter when y(t) (even for t < −t0) is being determined. Therefore, the system is NOT causal. t0

(x)

y (t ) =

dx

∫ x(λ)dλ + dt

−∞

(a) Linearity: For x1(t) and x2(t) applied as the inputs, the outputs are given by

y1 (t ) =

t0

∫ x (λ ) d λ + 1

−∞

y2 (t ) =

t0

∫ x (λ ) d λ + 2

−∞

dx1 , dt dx2 . dt

For x3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by

Solutions

y3 (t ) =

d (α x1 (t ) + β x2 (t ) ) dt

t0

∫ (α x ( λ ) + β x ( λ ) ) d λ + 1

57

2

−∞

 t0  t0 d ( x1 (t ) )  d ( x2 (t ) )  = α  ∫ x1 (λ )d λ +  + β  ∫ x2 (λ )d λ + , dt  dt   −∞  −∞ y1 ( t )

y2 ( t )

= α y1 (t ) + β y2 (t ) Therefore, the system is linear. (b) Time Invariance: For x1(t) and x2 (t ) = x1 (t − T ) applied as the inputs, the outputs are given by

x1 (t ) → y1 (t ) =

t0

∫ x (λ ) d λ + 1

−∞

x2 (t ) = x1 (t − T ) → y2 (t ) =

dx1 dt

t0

∫ x (λ ) d λ + 2

−∞

dx2 . dt

Substituting x2 (t ) = x1 (t − T ) we obtain,

y2 (t ) =

t0

∫ x (λ − T ) d λ + 1

−∞

By substituting λ´ = λ – T, we get y2 (t ) =

dx1 (t − T ) . dt

t0 −T

∫

x1 (λ ′)d λ ′ +

−∞

We also note that

y1 (t − T ) =

t0

∫ x (λ ) d λ + 1

−∞

dx1 (t − T ) . dt

dx1 (t − T ) , dt

implying that y1 (t − T ) ≠ y2 (t ) . The system is, therefore, NOT time invariant. (c) Stablity: Assume that the input is bounded |x(t)| ≤ M. Then, the output t0

y (t ) =

∫

dx(t ) x ( λ ) dλ + 2 ≤ dt

−∞

t0

∫ x ( λ ) dλ + 2

−∞

dx(t ) dt

is unbounded because of the integral which integrates x(t) from (−∞ ≤ t ≤ t0). Therefore, the system is NOT stable. (d) Causality: To solve the integral, the output y(t) always requires only the values of the input x(t) within the range (−∞ ≤ t ≤ t0) no matter when y(t) (even for t < −t0) is being determined. Therefore, the system is NOT causal. (xi)

d4y d3y d2y dy d 2x + 3 + 5 + 3 + y ( t ) = + 2 x(t ) + 1 dt dt 4 dt 3 dt 2 dt 2 (a) Linearity: For x1(t) applied as the input, the output y1(t) is given by

58

Chapter 2

d 4 y1 d 3 y1 d 2 y1 dy1 d 2 x1 y t + 3 + 5 + 3 + ( ) = + 2 x1 (t ) + 1 . 1 dt dt 4 dt 3 dt 2 dt 2

(S2.9.1)

For x2(t) applied as the input, the output y2(t) is given by d 4 y2 d 3 y2 d 2 y2 dy2 d 2 x2 + 3 + 5 + 3 + y ( t ) = + 2 x2 (t ) + 1 . 2 dt dt 4 dt 3 dt 2 dt 2

(S2.9.2)

For y3(t) = α x1(t) + β x2(t) applied as the input, the output y3(t) is given by d 4 y3 dt 4

+3

d 3 y3

+5

dt 3

d 2 y3 dt 2

+3

dy3 d 2 (αx1 (t ) + β x2 (t ) ) + y2 (t ) = + 2(αy1 (t ) + βy2 (t ) ) + 1 , dt dt 2

2 3 4   2   2 or, d y3 + 3 d y3 + 5 d y3 + 3 dy3 + y2 (t ) = α  d x1 + 2 x1 (t ) + 1 + β  d x2 + 2 x2 (t ) + 1 + [1 − α − β] . 2 2 2 3 4

dt

dt

dt

dt



 dt



 dt

Term I

Term II

Substituting the values of the derivative ( I and II) from Eqs. (S2.9.1) ans (2.9.2), we obtain

d 4 y3 dt 4

+3

d 3 y3 dt 3

+5

d 2 y3 dt 2

+3

d4y  dy3 d3y d2y dy + y2 (t ) = α  41 + 3 31 + 5 21 + 3 1 + y1 (t )  dt dt dt dt  dt  Term I

d4y  d3y d2y dy + β  42 + 3 32 + 5 22 + 3 2 + y2 (t ) + [1 − α − β] dt dt dt  dt  Term II

y3 (t ) ≠ αy1 (t ) + β y2 (t ) .

which implies that

The system is, therefore, NOT linear. Note that the dc term of (+ 1) on the right hand side of the differential equation contributes to the nonlinearity of the system (b) Time-invariance: The system is time-invariant. The proof is similar to Problem 2.1. (b) Time-invariance: For x1(t) and x2 (t ) = x1 (t − T ) applied as the inputs, the outputs are given by d 4 y1 d 3 y1 d 2 y1 dy1 d 2 x1 y t + 3 + 5 + 3 + ( ) = + 2 x1 (t ) + 1 1 dt dt 4 dt 3 dt 2 dt 2

(S2.9.3)

d 4 y2 d 3 y2 d 2 y2 dy2 d 2 x2 + 3 + 5 + 3 + y ( t ) = + 2 x2 (t ) + 1 . 2 dt dt 4 dt 3 dt 2 dt 2 Substituting x2 (t ) = x1 (t − T ) we obtain,

d 4 y2 d 3 y2 d 2 y2 dy2 d 2 x1 (t − T ) + 3 + 5 + 3 + y ( t ) = + 2 x1 (t − T ) + 1 . 2 dt 4 dt 3 dt 2 dt dt 2

(S2.9.4)

Substituting τ = t + T (which implies that dt = dτ) in Eq. (S2.9.3), we obtain

d 4 y1 (τ − T ) d 3 y1 (τ − T ) d 2 y1 (τ − T ) dy1 (τ − T ) d 2 x1 (τ − T ) + 3 + 5 + 3 + y ( τ − T ) = + 2 x1 (τ − T ) + 1 . 1 dτ 4 dτ 3 dτ 2 dτ dτ 2

Solutions

Or,

59

d 4 y1 (t − T ) d 3 y1 (t − T ) d 2 y1 (t − T ) dy1 (t − T ) d 2 x1 (t − T ) + 3 + 5 + 3 + y ( t − T ) = + 2 x1 (t − T ) + 1 . 1 dt 4 dt 3 dt 2 dt dt 2

Comparing with Eq. (S2.9.4), we obtain

y2 (t ) = y1 (t − T ) , proving that the system is time-invariant. (c) Stablity: The system is BIBO stable since a bounded input will always produce a bounded output. (d) Causality: Express Eq. (S2.2) as follows: t

τ

t

y (t ) = −3 ∫ y (α ) dα − 5 ∫ −∞

t

∫

−∞ −∞

t

+5∫

τ

∫

−∞ −∞

y (α )dα dτ − 3 ∫

τ

θ

∫∫

y (α )dα dτ dθ −

−∞ −∞ −∞

t

x (α )dα dτ + 2 ∫

τ

θ

∫∫

τ

φ

θ

∫ ∫ ∫ ∫ y(α )dα dτ dθ dφ

−∞ −∞ −∞ −∞

y (α )dα dτ dθ +

−∞ −∞ −∞

t

t

τ

θ

φ

∫ ∫ ∫ ∫ dα dτ dθ dφ

−∞ −∞ −∞ −∞

The output y(t) at t = t0 is given by t0

t0 τ

−∞

−∞ −∞

y (t ) t =t = −3 ∫ y (α )dα − 5 ∫ 0

t0 τ

t0 τ

θ

φ

∫ y(α )dα dτ − 3 ∫ ∫ ∫ y(α )dα dτ dθ − ∫ ∫ ∫ ∫ y(α )dα dτ dθ dφ −∞ −∞ −∞

t0 τ

t0 τ

−∞ −∞

−∞ −∞ −∞

+5∫

θ

θ

−∞ −∞ −∞ −∞

t0 τ

θ

φ

∫ x(α )dα dτ + 2 ∫ ∫ ∫ y(α )dα dτ dθ + ∫ ∫ ∫ ∫ dα dτ dθ dφ −∞ −∞ −∞ −∞

The system is causal since only the past values of the input x(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. ▌ Problem 2.10

(i)

y[k ] = ax[k ] + b

(a)

Linearity: Since x1[ k ] → ax1[ k ] + b = y1[ k ] x2 [ k ] → ax2 [ k ] + b = y1[k ]

(α x1[k ] + β x2 [k ]) → a (α x1[k ] + β x2 [k ]) + b ≠ α y1[k ] + β y2 [k ] the system is NOT a linear system. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[ k ] → ax1[ k ] + b = y1[ k ] x2 [ k ] = x1[ k − K ] → ax2 [k + 2] + b = y2 [ k ]

We note that

y2 [ k ] = ax2 [ k ] + b = ax1[k − K + 2] + b = y1[k − K ]

which implies that the system is time invariant. (c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Then, the output y[k ] = ax[k ] + b ≤ ax[k ] + b ≤ aM + b

is also bounded. Therefore, the system is BIBO stable.

60

Chapter 2

(d)

Causality: Since the output requires only the current value of input, the system is causal.

(ii)

y[k ] = 5 x[3k − 2]

(a)

Linearity: Since x1[ k ] → y1[k ] = 5 x1[3k − 2] x2 [k ] → y 2 [k ] = 5 x2 [3k − 2] (αx1[k ] + βx2 [k ]) → 5(αx1[3k − 2] + βx2 [3k − 2]) = αy1[k ] + βy2 [k ]

the system is a linear system. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[k ] → 5 x1[3k − 2] = y1[k ] x2 [k ] = x1[k − K ] → 5 x2 [3k − 2] x

2

[ k ] = x1 [ k − K ]

= 5 x1[3k − 2 − K ] = y 2 [k ]

y1[k − K ] = 5 x1[3(k − K ) − 2] ≠ y2 [k ]

We also note that

Therefore, the system is NOT time invariant. (c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Then, the output y[k ] = 5 x[3k − 2] = 5 x[3k − 2] ≤ 5M

is also bounded. Therefore, the system is BIBO stable. (d) Causality: When k ≥ 2, the output requires the future value of the input. Therefore, the system is NOT causal. (iii)

y[k ] = 2 x[ k ]

(a)

Linearity: Since x1[ k ] → 2 x1 [ k ] = y1[ k ] x2 [ k ] → 2 x2 [ k ] = y2 [ k ]

(α x1[k ] + β x2 [k ]) → 2α x [ k ]+ β x [ k ] = 2α x [ k ]2 β x [ k ] = ( 2 x [ k ] ) 1

2

1

2

1

α

( 2 ) = ( y [ k ]) ( y [k ]) x2 [ k ] β

α

1

≠ α y1[ k ] + β y2 [ k ]

the system is NOT a linear system. (b) Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[ k ]

→ 2 x1 [ k ] = y1[k ]

x2 [ k ] = x1[k − K ] → 2 x2 [ k ] = y2 [ k ]

Note that

y2 [ k ] = 2 x2 [ k ] = 2 x1 [ k − K ]

and

y1[ k − K ] = 2 x1 [ k ]

k =k −K

= 2 x1 [ k − K ] .

Since y2 [ k ] = y1[ k − K ] , the system is time invariant. (c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Then, the output

2

β

Solutions

y[k ] = 2 x[ k ] ≤ 2

x[ k ]

61

≤ 2M

is also bounded. Therefore, the system is BIBO stable. (d) Causality: Since the output requires only the current value of the input, therefore, the system is causal. (iv)

y[k ] =

k

∑ x[m]

m = −∞

(a)

Linearity: Since k

∑

x1[ k ] →

m =−∞

x2 [ k ] →

k

∑

m =−∞

x1[m] = y1[k ] x2 [ m] = y2 [ k ] k

k

m =−∞

m =−∞

(α x1[k ] + β x2 [k ]) → ∑ (α x1[m] + β x2 [m]) = α ∑

x1[ m] + β

k

∑

m =−∞

x2 [ m]

= α y1[ k ] + β y2 [ k ]

the system is a linear system. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[ k ] →

k

∑

m =−∞

x1[m] = y1[ k ]

x2 [ k ] = x1[k − K ] →

Note that

y2 [ k ] =

k

∑

m =−∞

x2 [ m] =

k

∑

m =−∞

k

∑

m =−∞

x2 [ m] = y2 [ k ]

x1[ m − K ] =

k −K

∑

m =−∞

x1[ m]= y1[ k − K ]

And hence the system is time invariant. (c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Then, the output y[k ] =

k

∑

x[m] ≤

m = −∞

k

∑ x[m]

m = −∞

may become unbounded as an infinite number of absolute values of x[k] are being added. Therefore, the system is NOT BIBO stable. (d) Causality: Since the output does not depend on the future values of the input, therefore, the system is causal. (v)

y[k ] =

k +2

∑ x[m] − 2 | x[k ] |

m=k −2

(a)

Linearity: Since

62

Chapter 2 k +2

∑ x1[m] − 2 x1[k ] = y1[k ]

x1[k ] → x2 [ k ] →

(αx1[k ] + βx2 [k ]) →

m=k −2 k +2

∑ x2 [ m ] − 2 x2 [ k ] = y 2 [ k ]

m=k −2 k +2

∑ (αx1[m] + βx2[m]) − 2 (αx1[k ] + βx2[k ])

m=k −2  k +2

  k +2  ≠ α x1[m] − 2 x1[k ]  + βα  x2 [ m ] − 2 x2 [ k ]       m=k −2   m=k −2 

∑

∑

the system is NOT a linear system. Note that the absolute term on the right hand side of the equation makes the system nonlinear. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[k ] →

k +2

∑

x1[m] − 2 x1[k ] m=k −2 k +2

x2 [k ] = x1[ k − K ] →

= y1[k ]

∑ x 2 [ m ] − 2 x2 [ k ] = y 2 [ k ]

m=k −2

Note that  k +2  y2 [k ] =  x 2 [ m ] − 2 x2 [ k ]  m = k − 2  x [ k ]= x [ k − K ] 2 1

∑

=

k +2

k −K +2

m=k −2

p=k − K −2

∑ x1[m − K ] − 2 x1[k − K ] = ∑ x1[ p] − 2 x1[k − K ]

which is the same as y1[k − K]. Therefore, the system is time invariant. (c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Then, the output y[ k ] =

k +2

∑

x[m] − 2 x[k ] ≤

m=k −2

k +2

∑ x[m] + 2 x[k ] ≤ 7M

m=k −2

is also bounded. Therefore, the system is BIBO stable. (d)

Causality: Since the output requires future values of the input, therefore, the system is NOT causal.

(vi)

y[k ] + 5 y[ k − 1] + 9 y[ k − 2] + 5 y[k − 3] + y[k − 4] = 2 x[k ] + 4 x[k − 1] + 2 x[k − 2]

(a)

Linearity: For x1[k] applied as the input, the output y1[k] is given by y1[k ] + 5 y1[ k − 1] + 9 y1[k − 2] + 5 y1[k − 3] + y1[k − 4] = 2 x1[k ] + 4 x1[k − 1] + 2 x1[ k − 2] .

For x2[k] applied as the input, the output y2[k] is given by

y2 [k ] + 5 y2 [k − 1] + 9 y2 [k − 2] + 5 y2 [k − 3] + y2 [k − 4] = 2 x2 [k ] + 4 x2 [k − 1] + 2 x2 [k − 2] . For x3[k] = α x1[k] + β x2[k] applied as the input, the output y3[k] is given by

Solutions

63

y3[k ] + 5 y3[k − 1] + 9 y3[k − 2] + 5 y3[k − 3] + y3[k − 4] = 2(αx1[k ] + β x2 [k ]) + 4(αx1[k − 1] + β x2 [ k − 1]) + 2(αx1[k − 2] + β x2 [k − 2]), which can be expressed as y3[k ] + 5 y3[k − 1] + 9 y3[k − 2] + 5 y3[k − 3] + y3[k − 4] = α(2 x1[k ] + 4 x1[k − 1] + 2 x1[k − 2]) + β(2 x1[k ] + 4 x1[k − 1] + 2 x1[k − 2]).

Substituting the value of the input , we get y3[k ] + 5 y3[k − 1] + 9 y3[k − 2] + 5 y3[k − 3] + y3[k − 4] = α( y1[k ] + 5 y1[k − 1] + 9 y1[ k − 2] + 5 y1[k − 3] + y1[k − 4]) + β( y2 [k ] + 5 y2 [k − 1] + 9 y2 [k − 2] + 5 y2 [k − 3] + y2 [k − 4]) or, y3[k ] + 5 y3[k − 1] + 9 y3[k − 2] + 5 y3[k − 3] + y3[k − 4] = (αy1[k ] + βy2 [k ]) + 5(αy1[ k − 1] + β y2 [ k − 1]) + 9(αy1[k − 2] + β y2 [k − 2]) + 5(αy1[k − 3] + β y2 [ k − 3]) + (αy1[k − 4] + β y2 [ k − 4]) , which implies that y3[k] = α y1[k] + β y2[k]. Therefore, the system is linear. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[k] → y1[k ] + 5 y1[k − 1] + 9 y1[k − 2] + 5 y1[k − 3] + y1[k − 4] = 2 x1[k ] + 4 x1[k − 1] + 2 x1[k − 2] x2[k] = x1[k − K] → y2 [k ] + 5 y2 [k − 1] + 9 y2 [ k − 2] + 5 y2 [ k − 3] + y2 [k − 4] = 2 x2 [k ] + 4 x2 [k − 1] + 2 x2 [k − 2] From the above equations, we can prove that y2[k] = y1[k − K] implying that the system is time invariant.

(c)

Stability: Assume that the input is bounded |x[k]| ≤ M. Taking the absolute value of both sides of the difference equation, we get y1[k ] + 5 y1[k − 1] + 9 y1[k − 2] + 5 y1[k − 3] + y1[k − 4] = 2 x1[k ] + 4 x1[ k − 1] + 2 x1[k − 2]

implying that

y1[k ] + 5 y1[k − 1] + 9 y1[ k − 2] + 5 y1[k − 3] + y1[k − 4] ≤ 8M .

Since sum of several output samples are bounded, y[k] itself must be bounded. Therefore, the system is BIBO stable. (d)

Causality: Since the output can be computed iteratively from y[k ] = (2 x[ k ] + 4 x[k − 1] + 2 x[k − 2]) − (5 y[k − 1] + 9 y[k − 2] + 5 y[k − 3] + y[k − 4]) ,

which requires only the previous values of the input, therefore, the system is causal. (vii) y[k ] = 0.5 x[6k − 2] + 0.5 x[6k + 2] (a)

Linearity: For x1[k] applied as the input, the output y1[k] is given by

y1[k ] = 0.5 x1[6k − 2] + 0.5 x1[6k + 2] . For x2[k] applied as the input, the output y2[k] is given by y2 [k ] = 0.5 x2 [6k − 2] + 0.5 x2 [6k + 2] .

64

Chapter 2

For x3[k] = α x1[k] + β x2[k] applied as the input, the output y3[k] is given by y3[k ] = 0.5(αx1[6k − 2] + βx2 [6k − 2]) + 0.5(αx1[6k + 2] + βx2 [6k + 2]) which can be expressed as y3[k ] = α(0.5 x1[6k − 2] + 0.5 x1[6k + 2]) + β(0.5 x2 [6k − 2] + 0.5 x2 [6k + 2]) , y1 [ k ]

y2 [k ]

implying that y3[k] = α y1[k] + β y2[k]. Therefore, the system is linear. (b)

Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[k ] → y1[k ] = 0.5 x1[6k − 2] + 0.5 x1[6k + 2] x2 [k ] = x1[k − K ] → 0.5 x2 [6k − 2] + 0.5 x2 [6k + 2] = y2 [k ]

The second equation implies that y2 [k ] = [0.5 x2 [6k − 2] + 0.5 x2 [6k + 2]]x [ k ]= x [ k − K ] 2 1 = [0.5 x1[6k − 2 − K ] + 0.5 x2 [6k + 2 − K ]]

We also note that

y1[ k − K ] = 0.5 x1[6k − 6 K − 2] + 0.5 x1[6k − 6 K + 2]

Since y2[k] ≠ y1[k − K], the system is NOT time invariant. (c)

Stability: Assuming that the input is bounded |x[k]| ≤ M, the output y1[k ] = 0.5 x1[6k − 2] + 0.5 x1[6k + 2] ≤ 0.5 x1[6k − 2] + 0.5 x1[6k − 2] ≤ M

is also bounded. Therefore, the system is BIBO stable. (d)

Causality: Since the output requires future values of the input, the system is NOT causal.

Problem 2.11

(i)

Using the linearity property 5 x(t ) → 5 y (t ) . The output waveform is drawn in Fig. S2.11(a).

(ii)

Using the time invariance property x(t − 1) → y (t − 1)

and

x(t + 1) → y (t + 1) .

Using the above result and the linearity property

[0.5 x(t − 1) + 0.5 x(t + 1)] → [0.5 y(t − 1) + 0.5 y(t + 1)] The output waveform is drawn in Fig. S2.11(b).

.

▌

Solutions

5x(t) → 5y(t) 5

−2

−1

1

t

2

(a) 0.5x(t−1) + 0.5x(t+1) → 0.5y(t−1) + 0.5y(t+1) 1 0.5 −2

−1

1

t

2

(b) x(t+1) − x(t−1) → y(t+1) − y(t−1) 1 1 −2

−1

t

2 −1

(c) dx (t ) dt

+ 3 x(t ) →

δ(t)

dy (t ) dt

4 2

−2

+ 3 y (t )

1

−1

2

t

(d) Fig. S2.11: Output waveforms for CT transformations in Problem 2.11. (iii) Using the time invariance property, we obtain x(t − 1) → y (t − 1) and

x(t + 1) → y (t + 1) .

Using the above result and the linearity property, we obtain

[ x(t + 1) − x(t − 1)] → [ y (t + 1) + y (t − 1)] . The output waveform is drawn in Fig. S2.11(c). (iv)

Since differentiation is a linear operation, dx(t ) dy (t ) → dt dt Using the above result and the linearity property

65

66

Chapter 2

 dx(t )   dy (t )   dt + 3x(t )  →  dt + 3 y (t ) .     The output waveform is drawn in Fig. S2.11(d).

▌

Problem 2.12

In of the DT impulse function δ[k], the output can be represented by y[k ] = 4δ[k + 2] − 2δ[k + 1] + 4δ[k ] + 2δ[k − 1] + 4δ[k − 2]. (i)

Using the linearity and time invariance properties, we obtain 4 x[k − 1] → 4 y[k − 1] = 16δ[k + 1] − 8δ[k ] + 16δ[k − 1] + 8δ[k − 2] + 16δ[k − 3]. The output waveform is drawn in Fig. S2.12(a).

(ii)

Using the time invariance property, we obtain x[k − 2] → y[k − 2] = 4δ[k ] − 2δ[k − 1] + 4δ[ k − 2] + 2δ[k − 3] + 4δ[k − 4] and

x[k + 2] → y[ k + 2] = 4δ[k + 4] − 2δ[k + 3] + 4δ[k + 2] + 2δ[k + 1] + 4δ[k ] .

Using the above result and the linearity property, we obtain

[0.5 x[k − 2] + 0.5 x[k + 2]] → [0.5 y[k − 2] + 0.5 y[k + 2]] or, [0.5 x[k − 2] + 0.5 x[k + 2]] → 2δ [k + 4] − δ [k + 3] + 2δ [k + 2] + δ [k + 1] +4δ [k ] − δ [k − 1] + 2δ [k − 2] + δ [k − 3] + 2δ [k − 4]. The output waveform is drawn in Fig. S2.12(b). (iii) Using the time invariance property, we obtain x[k + 1] → y[k + 1] = 4δ[k + 3] − 2δ[k + 2] + 4δ[k + 1] + 2δ[k ] + 4δ[k − 1] and

x[k − 1] → y[ k − 1] = 4δ[k + 1] − 2δ[k ] + 4δ[k − 1] + 2δ[k − 2] + 4δ[ k − 3] .

Using the above result and the linearity property, we obtain

[x[k − 1] − 2 x[k ] + x[k + 1]] → [ y[k − 1] − 2 y[k ] + y[k + 1]] or, [x[ k − 1] − 2 x[k ] + x[ k + 1]] → 4δ[k + 3] − 10δ[ k + 2] + 12δ[k + 1] − 8δ[k ] + 4δ[ k − 1] − 6δ[ k − 2] + 4δ[k − 3]. The output waveform is drawn in Fig. S2.12(c). (iv)

Using the time invariance property, we obtain x[−k ] → y[ −k ] = 4δ[ −k + 2] − 2δ[ −k + 1] + 4δ[− k ] + 2δ[−k − 1] + 4δ[− k − 2] . Using the property δ [−k + m] = δ [k − m] , the above equation reduces to x[−k ] → y[−k ] = 4δ[k − 2] − 2δ[k − 1] + 4δ[k ] + 2δ[k + 1] + 4δ[k + 2] The output waveform is drawn in Fig. S2.12(d).

▌

Solutions

0.5y[k – 2] + 0.5y[k + 2]

4y[k – 1]

−4

16

16

≈

≈

−2

67

16 8

≈

0

2

2

2 1

2

≈

4

≈

2 1

k

4

−4

−2

0

−1

2

4

k

−1

−8

(a)

(b) y[–k]

y[k – 1] – 2y[k ] + y[k + 1]

4

10 12 4

≈ −4

≈ ≈

−2

4

≈ 0

≈ −8

2

4 2

≈

4

4

≈ 4

k

−4

−2

0

2

4

k

−2

−6

(c)

(d)

Fig. S2.12: Output waveforms for DT transformations in Problem 2.12. Problem 2.13

(i)

The system is invertible with the inverse system given by x(t ) = 13 y (t − 2) .

(ii)

To calculate the inverse system, we differentiate the integral to get dy (t ) = x(t − 10) . dt The inverse system is obtained through two steps. Step 1 compute z(t) = dy/dt, while Step 2 computes x(t) from the relationship x(t) = z(t + 10).

(iii) The system y(t) = |x(t)| is not invertible as x(t) = ±a produces the same output y(t) = a. (iv)

If y(t) is differentiable then x(t) can always be calculated uniquely from the expression x(t ) =

dy (t ) + y (t ) dt

and the system is invertible. However, if y(t) is not differentiable (for example, it contains a discontinuity), then x(t) cannot always be calculated uniquely and the system is not invertible. (v)

System represented by y(t) = cos(2πx(t)) is not invertible as different values of x(t) = (θ + 2mπ), where m is an integer, produce the same output.

Problem 2.14

(i)

Not invertible as y[−1] is always 0 irrespective of the value of x[k].

(ii)

The input-output relationship can be expressed as:

68

Chapter 2 y[0] = x[2] y[ −1] = y[1] = x[2] + x[3] y[ −2] = y[2] = x[2] + x[3] + x[4] ............

It is observed that the output y[k ], − ∞ < ∞ depends on the x[k ], k ≥ 2 . From the output values, the value of x[k ], k ≥ 2 can be uniquely calculated. However, the input value x[k ], k < 2 cannot be calculated from y[k ] . Therefore the system is not invertible. (iii) The system is not invertible as x[k ] k = 0, ± 2, ± 4,... y[k ] =  elsewhere  0 All odd values of x[k] are lost and can not be recovered from y[k]. (iv)

System is invertible with the inverse system given by

x[k + 2] = y[k ] − (2 x[k + 1] − 6 x[k ] + 2 x[k − 1] + x[k − 2]) , Or, x[k ] = y[k − 2] − 2 x[k − 1] + 6 x[k − 2] − 2 x[k − 3] − x[k − 4] . (v)

System is invertible with the inverse system given by x[k ] = y[k ] + 2 y[k − 1] + y[k − 2] .

Problem 2.15

The proof can be made as follows.

x (t ) → y (t )

[ time invariance property ]

or, x (t + ∆t ) → y (t + ∆t )

x (t + ∆t ) − x (t ) y ( t + ∆t ) − y ( t ) → [linearity property ] ∆t ∆t x ( t + ∆t ) − x ( t ) y ( t + ∆t ) − y ( t ) or, lim → lim ∆t → 0 ∆ t → 0 ∆t ∆t dx dy → or, dt dt or,

Problem 2.16

The periodic signal xp(t) can be expressed in of x(t) as follows x p (t ) =

∞

∑ x(t − 2m),

m = integers .

m =−∞

Applying the linearity property, the output for the periodic signal xp(t) is given by x p (t ) =

which is shown in Fig. S2.16.

∞

∞

m =−∞

m =−∞

∑ x (t − 2m ) → ∑

y (t − 2 m ) = y p ( t )

Solutions

69

yp(t) 1

…

…

−6

−5

−4

−3

−2

−1

0

2

1

3

4

5

t

6

Fig. S2.16: Output yp(t) for the periodic signal xp(t) in Problem 2.16. Problem 2.17

The impulse response of the system is defined in the problem as follows:

δ (t ) → h(t ) or, h(t ) = y (t ) x ( t )=δ ( t ) . (i)

Substituting x(t) = δ(t), we obtain h(t ) = δ(t + 2) − 2δ(t ) + 2δ(t − 2) .

(ii)

Substituting x(t) = δ(t), we obtain

h (t ) =

t + to

∫

δ (τ − 4)dτ =

t −to

t + to − 4

∫

1 t − to − 4 < 0 < t + to − 4 otherwise 0

δ (λ )d λ = 

t −to − 4

1 4 − to < t < 4 + to = otherwise. 0

.

(iii) Substituting x(t) = δ(t), we obtain t

h (t ) =

∫

e −2( t −τ )δ (τ − 4)dτ =

−∞

∫

−∞

t

1

∫ δ (τ − 4)dτ = 0

Noting that

t

−∞

t

e −2( t −4)δ (τ − 4)dτ = e −2( t −4) ∫ δ (τ − 4)dτ −∞

t>4 = u (t − 4) , the impulse response is obtained as follows: otherwise

h(t ) = e −2( t −4)u(t − 4) . (iv)

Substituting x(t) = δ(t), we obtain ∞

h (t ) =

∫

f (T − τ )δ (t − τ )dτ =

−∞

∞

∫

−∞

∞

f (T − t )δ (t − τ )dτ = f (T − t ) ∫ δ (t − τ )dτ −∞

=1

= f (T − t ) Problem 2.18

The output h[k] of a DT LTI system to an unit impulse function δ[k] is shown in Fig. P2.18. Find the output for the following set of inputs. (i)

x[k ] = δ[k + 1] + δ[ k ] + δ[k − 1]

(ii)

x[k ] =

∞

∑ δ[k − 4m]

m = −∞

70

Chapter 2

(iii) x[k] = u[k] h[k] 1

1

−1

1

k

−2

Fig. P2.18: Output h[k] for input x[k] = δ[k] in Problem 2.18. Solution:

In of the DT impulse function δ[k], the impulse response h[k] is expressed as δ[k ] → h[k ] = δ[k + 1] − 2δ[k ] + δ[ k − 1]. (i)

Using the linearity and time invariance property

δ[k ] → h[k ] = δ[k + 1] − 2δ[k ] + δ[k − 1] δ[k + 1] → h[k + 1] = δ[k + 2] − 2δ[k + 1] + δ[k ] and δ[k − 1] → h[k − 1] = δ[ k ] − 2δ[k − 1] + δ[ k − 2] Using the linearity property δ[k + 1] + δ[k ] + δ[k − 1] → h[k + 1] + h[k ] + h[k − 1] or, δ[k + 1] + δ[k ] + δ[k − 1] → δ[k + 2] − δ[ k + 1] − δ[k − 1] + δ[k − 2]. (ii)

Using the linearity and time invariance property ∞

∞

m =−∞

m =−∞

∑ δ [k − 4m] → ∑ h[k − 4m] = y[k ]

which will be a periodic signal (with period K = 4). One period of the output signal y[k ] is given below.

 1 k = −1  −2 k = 0  y[ k ] =  k =1 1  0 k = 2. (iii) Recall that u[k ] =

∞

∑

δ[k − m] → y[k ] =

m =0

∞

∑ h[k − m] ,

m =0

which results in the output y[k ] = h[k ] + h[k − 1] + h[k − 2] + h[k − 3] + Substituting different values of k, we get

Solutions (k = −2):

y[−2] = h[−2] + h[−3] + h[−4] + h[−5] + … = 0.

(k = −1):

y[−1] = h[−1] + h[−2] + h[−3] + h[−4] + … = 1.

(k = 0):

y[0] = h[0] + h[−1] + h[−2] + h[−3] + … = −1.

(k = 1):

y[1] = h[1] + h[0] + h[−1] + h[−2] + … = 0.

By expanding for other values of k, it can be shown that y[k] = 0 for |k| ≥ 2. In other words,

y[k ] = δ [k + 1] − δ [k ] . Problem 2.19

(i)

Substituting x[k] = δ[k], we obtain y[k ] = δ[ k ] − 2δ[k − 1] + δ[k − 2] .

(ii)

Substituting x[k] = δ[k − 1] + δ[k + 1], we get y[k ] = (δ[k − 1] + δ[k + 1]) − 2(δ[k − 2] + δ[k ]) + (δ[k − 3] + δ[k − 1]) = δ[k − 3] − 2δ[k − 2] + 2δ[ k − 1] − 2δ[k ] + δ[k + 1]. x[k ] = 3δ[k + 3] + 2δ[k + 2] + δ[ k + 1] + δ[k − 1] + 2δ[k − 2] + 3δ[k − 3].

(iii) Express

Substituting the above value of x[k] in the difference equation, we get y[k ] = ( 3δ [k + 3] + 2δ [k + 2] + δ [k + 1] + −2 (

δ [k − 1] + 2δ [k − 2] + 3δ [k − 3]) δ [k − 2] + 2δ [k − 3] + 3δ [k − 4])

3δ [k + 2] + 2δ [k + 1] + δ [k ] +

+(

δ [k − 3] + 2δ [k − 4] + 3δ [k − 5])

3δ [k + 1] + 2δ [k ] + δ [k − 1] +

y[k ] = 3δ[k + 3] − 4δ[k + 2] + 2δ[k − 1] − 4δ[k − 4] + 3δ[k − 5].

or, Problem 2.20

(i)

Linearity: For x1[k] applied as the input, the output y1[k] is given by y1[k ] =

1 4 x1[k − m] . 5 m=0

∑

For x2[k] applied as the input, the output y2[k] is given by y2 [k ] =

1 4 x2 [ k − m ] 5 m=0

∑

For x3[k] = α x1[k] + β x2[k] applied as the input, the output y3[k] is given by y3 [ k ] =

4 4 1 4 (αx1[k − m] + βx2 [k − m]) = α × 1 x1[k − m] + β × 1 x2 [k − m] 5 m=0 5 m=0 5 m =0

∑

∑

∑

which implies that y3[k] = α y1[k] + β y2[k]. Therefore, the system is linear. Time Invariance: For inputs x1[k] and x2[k] = x1[k − K], the outputs are given by x1[k ] → y1[k ] =

1 4 x1[k − m] 5 m=0

∑

71

72

Chapter 2

x2 [ k ] = x1[k − K ] → y2 [k ] =

and

1 4 1 4 x2 [ k − m ] = x1[ k − K − m] . 5 m =0 5 m=0

∑

∑

From the above equations, it is clear that y2[k] = y1[k − K] implying that the system is time invariant. (ii)

The impulse response of the system is given by h[k ] =

1 4 ∑ δ [k − m] = 0.2 (δ [k ] + δ [k − 1] + δ [k − 2] + δ [k − 3] + δ [k − 4]) 5 m =0

0.2 =  0

k = 0,1, 2,3, 4 otherwise

(iii) By expressing u[k ] =

∞

∑

δ[k − m] → y[k ] =

m =0

∞

∑ h[k − m] ,

m =0

the output y[k] to the unit step function is given by

y[k ] = h[k ] + h[k − 1] + h[k − 2] + h[k − 3] + h[k − 4] + h[k − 5] + Substituting different values of k, we get

(iv)

(k ≤ −1):

y[k] = 0.

(k = 0):

y[0] = h[0] + h[−1] + h[−2] + h[−3] + h[−4] + h[−5] … = 0.2.

(k = 1):

y[1] = h[1] + h[0] + h[−1] + h[−2] + h[−3] + h[−4] + … = 0.4.

(k = 2):

y[2] = h[2] + h[1] + h[0] + h[−1] + h[−2] + h[−3] +… = 0.6.

(k = 3):

y[3] = h[3] + h[2] + h[1] + h[0] + h[−1] + h[−2] + … = 0.8.

(k ≥ 4):

y[k] = 1

Using the linearity and time invariance property

δ [k ] = ( u[k ] − u[k − 1]) → h[k ] = ( y[k ] − y[k − 1]) which leads to

 0 k ≤ −1  0 k ≤ 0 0.2 k = 0 0.2 k = 1   k ≤ −1  0 0.4 k = 1 0.4 k = 2  − = 0.2 0 ≤ k ≤ 4 h[k ] =  = = 0.6 k 2 0.6 k 3    0 k ≥ 5.  0.8 k = 3  0.8 k = 4    1.0 k ≥ 4 1.0 k ≥ 5 y[k ]

y [ k −1]

Note that the above value of the impulse response is the same as the value obtained in part (ii). ▌ Problem 2.21

Solutions

73

(i)

The linearity and time invariance properties can be proved directly from the definitions. Instead, we use the property of linear, constant coefficient finite difference equations, which always represent linear and time invariant systems.

(ii)

Series Configuration: Denoting the output of system S1 by w[k], we obtain w[k ] = x[ k ] − 2 x[k − 1] + x[ k − 2] . For system S2, w[k] is the input and y[k] is the output. The output in given by y[k ] = w[k ] + w[k − 1] − 2 w[k − 2] . Substituting the value of w[k] from the earlier equation, we obtain y[k ] = (x[k ] − 2 x[k − 1] + x[k − 2]) + (x[k − 1] − 2 x[k − 2] + x[k − 3]) − 2(x[k − 2] − 2 x[k − 3] + x[k − 4])

which reduces to y[k ] = x[k ] − x[ k − 1] − 3x[ k − 2] + 5 x[k − 3] − 2 x[ k − 4] . (iii) Parallel Configuration: Denoting the output of system S1 by w1[k], we obtain

w1[k ] = x[k ] − 2 x[k − 1] + x[k − 2] . For system S2, x[k] is the input and w2[k] is the output. The output is given by w2 [k ] = x[ k ] + x[k − 1] − 2 x[k − 2] . The overall output of the parallel configuration is given by

y[k ] = w1 [k ] + w2 [k ] = ( x[k ] − 2 x[k − 1] + x[k − 2]) + (x[k ] + x[k − 1] − 2 x[k − 2]) , which reduces to y[k ] = 2 x[k ] − x[ k − 1] − x[k − 2] . (iv)

Since both series and parallel configurations are represented by linear, constant coefficient finite difference equations, both systems are linear and time invariant. ▌

74

Chapter 2

Additional work: Supplement to Problem 2.4 How do we demodulate the AM signal?

The modulated signal can be demodulated as follows: Multiplying the modulated signal s (t ) = A[1 + km(t )] cos(2πf c t ) with cos(2πfct), we get s (t ) = A[1 + km(t )] cos 2 (2πf c t ) =

1 2

A[1 + km(t )]

Low Frequency Component

+

1 2

A[1 + km(t )] cos(4πf c t ) . High Frequency Component

By using a low filter, the high frequency component can be filtered out. The low frequency component can be used be extract the information bearing signal m(t).

▌

Supplement to Problem 2.9 Determine if the systems are memoryless and invertible. If invertible, find the inverse system.

(i)

y (t ) = x (t − 2)

Also, note that the system requires past memory and is not memoryless. Further, the system is invertible as the input can be reconstructed using the relationship x(t ) = y (t + 2) . (ii)

y (t ) = x(2t − 5) Since all noncausal systems must have memory, the system is NOT memoryless. Furthermore, the system is invertible with the inverse system given by

y (t ) = x(2t − 5) ⇒ x(t ) = y ( t +25 ) = y (0.5t + 2.5) . (iii)

y (t ) = x(2t ) − 5 Since all noncausal systems must have memory, the system is NOT memoryless. Furthermore the system is invertible with the inverse system given by

y (t ) = x(2t ) − 5 ⇒ x(2t ) = y (t ) + 5 ⇒ x(t ) = y (0.5t ) + 5 . (iv)

y (t ) = tx (t + 10)

Since all noncausal systems must have memory, therefore, the system is NOT memoryless. Futhermore, the system is NOT invertible as its inverse

x (t ) =

y ( t −10) t −10

,

is not defined for t = 10. Therefore, the system is NOT invertible. (v)

 2 y (t ) = 2u ( x ( t ) ) =   0

x (t ) ≥ 0 x (t ) < 0

Also note that the system is memoryless since the output at any time instant does not depend on the past or future values of the input. The system is not invertible since it is not possible to calculate the input x(t) uniquely from y(t). The output is always 0 for x(t) < 0. Likewise, the output is always 2 for x(t) > 0. (vi)

 0 y (t ) =   x(t ) − x(t − 5)

t<0 t≥0

= [ x(t ) − x(t − 5) ] u (t )

Solutions

75

The system is NOT memoryless since it requires past values of the input to compute the current value of the output. For (t < 0), the output is always 0. Hence, it is not possible to calculate the input x(t) uniquely from y(t) for (t < 0). Therefore, the system is NOT invertible. (vii) y (t ) = 7 x 2 (t ) + 5 x(t ) + 3 The system is also memoryless since only the current value of the input is required to calculate the output. The system is NOT invertible as the inverse system given by x(t ) =

− 5 ± 25 − 28(3 − y (t ) 14

produces two possible values for inverting each value of y(t). (viii) y(t) = sgn(x(t)) The system is also memoryless since only the current value of the input is required to calculate the output. The system is NOT invertible because it is not possible to determine x(t) from y(t). All positive values of x(t) produce the same output of +1, while all negative values of x(t) produce the same output of −1. t0

(ix)

∫ x(λ)dλ + 2 x(t )

y (t ) =

−t0

The system is NOT memoryless as it is not causal. The system is also invertible. t0

(x)

dx

∫ x(λ)dλ + dt

y (t ) =

−∞

The system is NOT memoryless as it is not causal. The system is invertible. (xi)

d4y d3y d2y dy d 2x + 3 + 5 + 3 + y ( t ) = + 2 x(t ) + 1 dt dt 4 dt 3 dt 2 dt 2 The output y(t) at t = t0 is given by t0

t0 τ

−∞

−∞ −∞

y (t ) t =t = −3 ∫ y (α )dα − 5 ∫ 0

t0 τ

+5∫

∫

−∞ −∞

∫

t0 τ

y (α ) dα dτ − 3 ∫

θ

∫∫

y (α )dα dτ dθ −

−∞ −∞ −∞

t0 τ

x(α )dα dτ + 2 ∫

θ

∫∫

y (α )dα dτ dθ +

−∞ −∞ −∞

t0 τ

θ

φ

∫ ∫ ∫ ∫ y(α )dα dτ dθ dφ

−∞ −∞ −∞ −∞ t0 τ

θ

φ

∫ ∫ ∫ ∫ dα dτ dθ dφ

−∞ −∞ −∞ −∞

The system has memory since the past values of the input x(t), for −∞ ≤ t ≤ t0, are needed to calculate the output y(t) at t = t0. Invertiblity: The system is invertible.

▌

Supplement to Problem 2.10 Determine if the systems are memoryless and invertible.

(i)

y[k ] = ax[k ] + b Since the output requires only the current value of input, therefore, the system is both causal and memoryless.

76

Chapter 2 Further, the system is invertible with the inverse system given by x[k ] =

(ii)

1 a

( y[k ] + b ) .

y[k ] = 5 x[3k − 2] The system is NOT memoryless as it is not causal. Further, the system is invertible with the inverse system given by x[k ] =

(iii)

1 5

y

[k +3 2 ] .

y[k ] = 2 x[ k ] The system is memoryless as well as causal since the output does not depend on the past or future values of the input. Furthermore, the system is invertible with the inverse system given by

x[k ] = log 2 (x[k ]) . (iv)

y[k ] =

k

∑ x[m]

m = −∞

The system is NOT memoryless as the output requires past values of the input. Furthermore, the system is invertible with the inverse system given by x[k ] = y[k ] − y[k − 1] . (v)

y[k ] =

k +2

∑ x[m] − 2 | x[k ] |

m=k −2

The system is NOT memoryless as it is not causal. Further, the system is NOT invertible because of the |x[k]| term. (vi)

y[k ] + 5 y[ k − 1] + 9 y[ k − 2] + 5 y[k − 3] + y[k − 4] = 2 x[k ] + 4 x[k − 1] + 2 x[k − 2] The system is NOT memoryless as it requires some past values of the input. Furthermore, the system is invertible.

(vii) y[k ] = 0.5 x[6k − 2] + 0.5 x[6k + 2] The system is NOT memoryless as it is not causal. Furthermore, the system is invertible.

▌