Continuous and Discrete Time Signals and Systems (Mandal & Asif) Solutions - Chap02

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 terms 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 terms 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 lowpass 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. (vi)

 0 y (t ) =   x(t ) − x(t − 5)

t