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Chapter 2

Continuous-time Systems

1

Full file at http://testbank360.eu/solution-manual-signals-and-systems-using-matlab-1st-edition-luis-chaparro

Chaparro — Signals and Systems using MATLAB 2.2

Pr. 2.1 (a) The output voltage when the switch closes at t = 0 is

vo(t) = −R(t)u(t)

= −(1 + 0.5 cos(20πt))u(t)

The initial value of the voltage is v(0) = −1.5.(b) If the switch closes at t = 50 msec, the output voltage is

vo(t) = −R(t)u(t− 50× 10−3)

=

−(1 + 0.5 cos(20πt)) t ≥ 50× 10−3

0 otherwise

with an initial value of vo(50× 10−3) = −0.5.(c) The initial values are different, and v0(t) 6= vo(t− 50× 10−3) so the system is time varying.The following script is used for the plotting in (a) and (b)

%% Pr 2.1

t=0:0.01:0.2;M=length(t);

v0=-(1+0.5*cos(20*pi*t));

t1=0:0.01:0.05;

N=length(t1);

v01=[zeros(1,N) v0(N+1:M)];

figure(1)

subplot(211)

plot(t,v0); grid; axis([0 max(t) -1.7 0.2]); ylabel(’v_0(t)’)

subplot(212)

plot(t,v01); grid;axis([0 max(t) -1.7 0.2]); xlabel(’t’); ylabel(’v_01(t)’)

Copyright 2010, Elsevier, Inc. All rights reserved.



0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

−1.5

−1

−0.5

0

v 0(t)

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

−1.5

−1

−0.5

0

t

v 01(t

)

Figure 2.1:




Pr. 2.2 (a) The zener diode clips any signal that exceeds a set threshold, in this case 0.5, thus if we have asinusoid with amplitude 0.3, less than the threshold, the output of the zerner diode would be the same as theinput. When the amplitude of the input is 1 the output is a sinusoid clipped at 0.5, but when it is 0.3 is not aclipped sinuosoid. Thus, the system is non-linear as scaling does not hold.(b) The zener diode would do the same thresholding to a signal x(t) or to a shifted version of it x(t− τ), so

the system is time invariant.

%% Pr 2.2

t=0:0.001:4;

vs=cos(pi*t); x=vs;

% vs=0.3*cos(pi*t);x=vs; % for a second input get rid of %

N=length(t);

for k=1:N,

if abs(vs(k))>=0.5,

if vs(k)<0,

vs(k)=-0.5;

else

vs(k)=0.5;

end

end

end

y=vs;

figure(2)

plot(t,x); grid; hold on

plot(t,y,’or’); axis([0 4 -1.1 1.1]); legend(’x(t)’,’y(t)’); xlabel(’t’)

hold off

0 0.5 1 1.5 2 2.5 3 3.5 4

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

t

x(t)y(t)

Figure 2.2: Input x(t) and output y(t) of the zener diode.




Pr. 2.3 (a) Letting x(t) = δ(t) the impulse response is

h(t) =1

T

∫ t+T/2

t−T/2δ(τ)dτ

=1

T

∫ t

t−T/2δ(τ)dτ +

1

T

∫ t+T/2

t

δ(τ)dτ

If t > 0, and t− T/2 < 0 the first integral includes 0, while the second does not. Thus

h(t) =1

T

∫ t

t−T/2δ(τ)dτ + 0 =

1

Tt > 0 and t− T/2 < 0, or 0 < t < T/2

Likewise when t < 0 then t − T/2 < −T/2 and t + T/2 < T/2 the reverse of the previous case happensand so

h(t) = 0 +1

T

∫ t+T/2

t

δ(τ)dτ =1

Tt < 0 and t+ T/2 > 0, or − T/2 < t < 0

so thath(t) =

1

T[u(t+ T/2)− u(t− T/2)]

indicating that the system is non-causal as h(t) 6= 0 for t < 0.(b) If x(t) = u(t) then the output of the averager is

y(t) =1

T

∫ t+T/2

t−T/2u(τ)dτ

If t + T/2 < 0 then y(t) = 0 since the argument of the unit step signal is negative. If t + T/2 ≥ 0 andt− T/2 < 0 then

y(t) =

∫ t+T/2

0

u(τ)dτ =1

T(t+ T/2)

and finally when t− T/2 ≥ 0 then

y(t) =1

T

∫ t+T/2

t−T/2u(τ)dτ = 1

The unit-step response of the noncausal averager is

y(t) =

0 t < −T/21T (t+ T/2) −T/2 ≤ t < T/2

1 t ≥ T/2




Pr. 2.4 (a) The system is LTI since the input x(t) and the output y(t) are related by a convolution integralwith h(t− τ) = e−(t−τ)u(t− τ) or h(t) = e−tu(t).Another way: to show that the system is linear let the input be x1(t) + x2(t), and x1(t) and x2(t) have asoutputs

yi(t) =

∫ t

0

e−(t−τ)xi(τ)dτ i = 1, 2

The output for x1(t) + x2(t) is∫ t

0

e−(t−τ)(x1(τ) + x2(τ))dτ = y1(t) + y2(t)

To show the time invariance let the input be x(t− t0), its output will be∫ t

0

e−(t−τ)x(τ − t0)dτ =

∫ 0

−t0e−((t−t0)−µ)x(µ)dµ+

∫ t−t0

0

e−((t−t0)−µ)x(µ)dµ

=

∫ t−t0

0

e−((t−t0)−µ)x(µ)dµ = y(t− t0)

by letting µ = τ − t0 and using the causality of the input. The system is then TI.Finally the impulse response is found by letting x(t) = δ(t) so that the output is

h(t) =

∫ t

0

e−(t−τ)δ(τ)dτ =

∫ t

0

e−(t−0)δ(τ)dτ =

e−t × 1 = e−t t ≥ 0

0 otherwise

(b) Yes, this system is causal as the output y(t) depends on present and past values of the input.(c) Letting x(t) = u(t), the unit-step response is

s(t) =

∫ t

0

e−t+τu(τ)dτ = e−t∫ t

0

eτdτ = 1− e−t

for t ≥ 0 and zero otherwise. The impulse response as indicated before is h(t) = ds(t)/dt = e−tu(t).The BIBO stability of the system is then determined by checking whether the impulse response is absolutelyintegrable or not, ∫ ∞

−∞|h(t)|dt =

∫ ∞0

e−tdt = −e−t |∞0 = 1

so yes it is BIBO stable.(d) Using superposition, the response to the pulse x1(t) = u(t)− u(t− 1) would be

y1(t) = y(t)− y(t− 1) = (1− e−t)u(t)− (1− e−(t−1))u(t− 1)

which starts at zero, grows to a maximum of 1− e−1 at t = 1 and goes down to zero as t→∞.




Pr. 2.5 (a) No, the diode is non-linear. For v(t) > 0 doubling the voltage does not double the current i(t)because the voltage-current relation is not a 45o line.(b) To get the current to be close to zero when the voltage is negative we would need that Is → 0 becauseeqv(t)/kT −1 ≤ 0. If the constant q/kT →∞ the current would be very large for small values of the voltage.The ideal diode is an approximation to the p-n diode with the voltage-current plot given by the script below.The ideal diode is non-linear, the voltage-current plot is not a line of slope 45o. The following is the plot forthe given values of Is and the constant q/kT .

v=-0.1:0.01:1;

Is=0.0001;kqT=0.026;

i=Is*(exp(v/kqT)-1);

figure(1)

plot(v,i);grid

axis([-0.1 1.2 -5 1.1*max(i)])

0 0.2 0.4 0.6 0.8 1 1.20

1

2

3

4

5

x 1012

Figure 2.3: i-v response of the p-n diode.

(c) This is a half-wave rectifier used to obtain dc sources. The output is the positive part of the sinusoid andzero whenever the sinusoid is negative.

0 0.5 1 1.5 2 2.5 3 3.5 4

0

0.2

0.4

0.6

0.8

1

t

|x(t

)|

Figure 2.4: Output of the half-wave rectifier.




Pr. 2.6 (a) The output is given by

vc(t) =

∫ t

0

is(t′)dt′ + vc(0)

and for zero initial condition, it is linear. If is1(t) = u(t) − u(t − 1) and vc(0) = −1 the voltage in thecapacitor is given by

vc1(t) = −1 + r(t)− r(t− 1)

If we double the input i.e., is2(t) = 2[u(t)−u(t−1)] with the same initial condition, the output is then givenby

vc2(t) = −1 + 2r(t)− 2r(t− 1)

which is not 2vc1(t) so the system is non-linear.(b) The output is

iL(t) =

∫ t

0

vs(t′)dt′ + iL(0)

the system is linear if the initial current iL(0) = 0.For vs1 = u(t)− u(t− 1) and vs2 = 2vs1(t), the output currents are

iL1(t) = −1 + r(t)− r(t− 1)

iL2(t) = −1 + 2r(t)− 2r(t− 1)

where iL2(t) 6= 2iL1(t) indicating the system is not linear.




Pr. 2.7 (a) The charge isq(t) = C(t)v(t)

so thati(t) =

dq(t)

dt= C(t)

dv(t)

dt+ v(t)

dC(t)

dt

(b) If C(t) = 1 + cos(2πt) and v(t) = cos(2πt), the current is

i1(t) = C(t)dv(t)

dt+ v(t)

dC(t)

dt= (1 + cos(2πt))(−2π sin(2πt))− cos(2πt)(2π sin(2πt))

= −2π sin(2πt)[1 + 2 cos(2πt)]

(c) When the input isv(t− 0.25) = cos(2π(t− 1/4)) = sin(2πt)

the output current is

i2(t) = C(t)dv(t− 0.25)

dt+ v(t− 0.25)

dC(t)

dt

= (1 + cos(2πt))(2π cos(2πt))− 2π sin2(2πt)

= 2π cos(2πt) + 2π[cos2(2πt)− sin2(2πt)]

which is noti1(t− 0.25) = 2π cos(2πt)[1 + sin(2πt)]

so the system is time varying.




Pr. 2.8 The input to all the systems is x(t) = cos(t),−∞ < t <∞(1) The system is non-linear, as the output

y(t) = cos2(t) = 0.5(1 + cos(2t))

has frequency components of frequencies 0 and 2 (rad/sec) which are not in the input.(2) The output is

y(t) = 0.5 cos(t) + 0.5 cos(t− 1)

having the same frequencies as the input so it is LTI.(3) The output

y(t) = cos(t)u(t)

is not LTI. This is not a periodic signal, and it has frequencies different from the one at the input due to themultiplication by the u(t).(4) The output is

y(t) = 0.5 sin(τ)|tt−2 = 0.5 sin(t)− 0.5 sin(t− 2)

having the same frequency as the input so it is LTI.




Pr. 2.9 (1) The output due to x1(t) = u(t) is

y1(t) = cos(πt)u(t)

and the output due to x2(t) = x1(t− 1) = u(t− 1) is

y2(t) = cos(πt)u(t− 1) 6= y1(t− 1) = cos(π(t− 1))u(t− 1)

so the system is TV.(2) The output due to x1(t) = u(t) is

y1(t) = u(t)− u(t− 2)

while the output due to x2(t) = x1(t− 1) = u(t− 1) is

y2(t) = u(t− 1)− u(t− 2) 6= y1(t− 1) = u(t− 1)− u(t− 3)

so the system is TV.(3) The output due to x1(t) = u(t) is

y1(t) = 0.5[u(t) + u(t− 1)]

and for x2(t) = u(t− 1) it is

y2(t) = 0.5[u(t− 1) + u(t− 2)] = y1(t− 1)

so the system is TI.




Pr. 2.10 (a) Since f(t) is not a constant, the system is a modulator thus linear but time varying. Linearity isclearly satisfied. If x(t) 6= 0 is the input and we shift it to get as input x(t − 11) the corresponding outputis zero different from y(t − 1). Thus the system is time varying. Since y(t) depends on x(t) the system iscausal. For x(t) bounded, i.e., |x(t)| < M < ∞, the output is also bounded, |y(t)| < M |f(t)| < ∞ so thesystem is BIBO stable.(b) The modulated signal is

x(t)f(t) = 16[cos((π/2 + 6π/7)t) + cos((6π/7− π/2)t) = 16(cos(19πt/14) + cos(5πt/14)

with periods of T0 = 28/19 and T1 = 28/5 for the two components. The ratio

T0T1

=5

9

i.e., it is rational so the modulated signal is periodic of period 5T1 = 19T0 = 28, which is easily verified.The frequencies at the output are not present at the input so the system is linear but not time–invariant ( f(t)

is a function of t).(c) If x(t) = u(t), the modulated signal is y(t) = u(t) − u(t − 2), and if we shift the input so that it isx(t − 3) = u(t − 3) the corresponding output is u(t − 3)[u(t) − u(t − 2)] = 0 different from the previousoutput shifted by 3, therefore the system is time-varying.




Pr. 2.11 (a) If the initial condition y(0) 6= 0 the system is not LTI as it does not depend linearly on x(t), e.g.,if we double(b) To find the differential equation let y(0) = 0, so that

y(t) = 2e−2t∫ t

0

e2τx(τ)dτ

and its derivative is

dy(t)

dt= −4e−2t

∫ t

0

e2τx(τ)dτ + 2e−2te2tx(t) = −2y(t) + 2x(t)

giving the differential equationdy(t)

dt+ 2y(t) = 2x(t)

(c) Finding the integral after replacing x(τ) = u(τ) we have that

y(t) = y(0)e−2t + 2e−2te2τ

2

∣∣t0 = y(0)e−t + (1− e−2t)

so that in the steady state, i.e., when t→∞, the output is 1, independent of the value of the initial condition.(d) From the given input-output equation, letting x(t) = δ(t) and y(0) = 0 the output is the impulse responseof the system

h(t) = 2

∫ t

0

e−2(t−τ)δ(τ)dτ = 2e−2t∫ t

0

δ(τ)dτ = 2e−2tu(t)

Computing the convolution integral as

y(t) =

∫ ∞−∞

x(t− τ)h(τ)dτ

it can be obtained graphically by reflecting the input x(t) = u(t) and shifting it linearly from −∞ to∞ toget

• for t < 0 then y(t) = 0

• for t ≥ 0

y(t) =

∫ t

0

2e−2τdτ = 1− e−t

and so in the steady state the output goes to 1.(e) The zero-input response is

yzi(t) = y(0)e−2tu(t)

and it is bounded for any finite value of the initial condition y(0), in particular for y(0) = 1, therefore thesystem that depends on the initial condition is BIBO.




Pr. 2.12 (a) The y(t)-x(t) relation is a line through the origin between −10 to 10 and a constant before andafter that. The system is non-linear, for instance if x(t) = 7 the output is y(t) = 700 but if we double theinput, the output is not 2y(t) = 1400 but 1000.(c) If the inputs is always between −10 and 10 the system behaves like a linear system.(d) Whenever the input goes below −10 or above 10 the output is −1000 and 1000, otherwise the output is2000 cos(2πt)u(t).

−2 −1 0 1 2 3 4

−1000

−800

−600

−400

−200

0

200

400

600

800

1000

t

x(t)y(t)

Figure 2.5: Input and output of amplifier.

(e) If the input is delayed by 2 the clipping will still occur, simply at a later time. So the system is timeinvariant.




Pr. 2.13 (a) The output is the sum of two modulated signals, and as such it is time-varying as each modulatoris time-varying.(b) If we let m1(t) = m2(t) = m(t), the output is equal to the sum of two modulators so it is linear since themodulators are liner. Also

s(t) = m(t)(cos(Ωct) + sin(Ωct)) =√

2m(t) cos(Ωct− π/4)

i.e., one modulator with a phase. This system is known to be linear.




Pr. 2.14 This averager is a LTI system(a) The average signal for T = 1 and x(t) = u(t) is

y(t) =

∫ t

t−1u(τ)dτ =

0 t < 0

∫ t0dτ = t 0 ≤ t < 1

∫ tt−1 dτ = 1 t ≥ 1

Using the LTI of the system, the response to x1(t) = x(t)− x(t− 1) = u(t)− u(t− 1) is then

y1(t) = y(t)− y(t− 1) =

t 0 ≤ t ≤ 1

1− (t− 1) = 2− t 1 ≤ t ≤ 2

0 otherwise

which is a triangular signal.(b) The implementation of the averager can be seen as the sliding, from left to right, of a rectangular windowof width T and amplitude 1, and adding the values of the signal in that window to get the output. If we letT = T0, the period of the sinusoid x(t) = cos(2πt/T0)u(t) after zero, the output of the averager is zero fort ≤ 0. There is then a transient for 0 ≤ t ≤ T0, and the output will be zero for t ≥ t0, as the average of thesinusoids is zero.(c) The following script is used to implement this convolution in (a) and (b).

N=1000; Ts=1/N;

tt=0:Ts:2; x=cos(2*pi*tt);

h=[ones(1,N) zeros(1,N+1)];

y=conv(x,h)*Ts;t=0:Ts:(length(y)-1)*Ts;

figure(5)

subplot(211)

plot(tt,x);hold on;plot(tt,h,’or’); legend(’x(t)’,’h(t)’); axis([0 2 -1.1 1.1]); grid

subplot(212)

plot(t(1:length(t)/2),y(1:length(t)/2));grid

axis([0 2 1.1*min(y) 1.1*max(y)]); ylabel(’x*h’); xlabel(’t’)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2−1

−0.5

0

0.5

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

x*h

t

x(t)h(t)

Figure 2.6: Sinusoid and impulse response (top), result of convolution.




Pr. 2.15(a)

+x(t) y(t)

!1

!N!1

!N

!–delay

!–delay

!–delay

...

y(t! (N ! 1)!)

y(t!N!)

y(t! !)

Figure 2.7: Recursive model for echo system.

(b)The input/output equation for the echo system is

y(t) = x(t) + 0.1y(t− 1)

To check the LTI we need an explicit expression of the output in terms of the input. This can be obtained byrecursively replacing y(t−k), k = 1, 2, · · · in the right by the a delayed version of the input/output equation,i.e.,

y(t) = x(t) + 0.1y(t− 1) = x(t) + 0.1x(t− 1) + 0.01x(t− 2) · · ·

=

∞∑k=0

(0.1)kx(t− k)

Using this expression we have that if the input is x(t) = x1(t) +x2(t), with corresponding outputs y1(t) andy2(t), then

∞∑k=0

(0.1)k[x1(t− k) + x2(t− k)] = y1(t) + y2(t)

so the echo system is linear. If we shift the input to x(t− n0) then the output will be

∞∑k=0

(0.1)kx(t− n0 − k) = y(t− n0)

i.e., time-invariant. The system is then overall LTI.(c) The non-recursive model can be written as

z(t) =

M∑k=0

βkx(t− kτ) β0 = 1




x(t) !–delay !–delay !–delay· · ·

x(t! !) x(t! (M ! 1)!) x(t!M!)

!1 !2 !M

z(t)

+

Figure 2.8: Non-recursive model for echo system.

We have a finite representation of the output in terms of the inputs, attenuated and shifted in time. The systemresembles that in (b), and as in there it can be shown to be LTI.




Pr. 2.16 (a) The system with the sinc impulse response is non-causal because h(t) 6= 0 for t < 0. As such anideal low-pass filter cannot be used for real-time processing as it would require future inputs, not available inreal-time. This can be seen by looking at the convolution with a causal signal x(t):

y(t) =

∫ ∞0

x(τ)h(t− τ)dτ

=

∫ t

0

x(τ)h(t− τ)dτ +

∫ ∞t

x(τ)h(t− τ)dτ

where the upper limit of the top integral is due to h(τ) having an infinite support, and the lower limit becausex(t) is causal. The second bottom integral shows that the output depends on future values of the input.(b) For the low-pass filter to be BIBO stable the impulse response h(t) must be absolutely integrable, i.e.,∫ ∞

−∞| sin(t)/t|dt <∞

To approximtely compute this integral we use the following script. The integral is bounded, so the filter isBIBO stable.

clear all; clf

syms x t x1

x=abs(sinc(t/pi));

for k=1:30,

x1=int(x,t,0,k);

xx(k)=subs(2*x1);

end

n=1:30;

figure(1)

subplot(211)

ezplot(x,[-30,30]); axis([-30 30 -0.1 1.2]);grid

subplot(212)

stem(n,xx); axis([1 30 0 8]);grid

−30 −20 −10 0 10 20 30

0

0.2

0.4

0.6

0.8

1

t

abs(sin(t))/abs(t)

0 5 10 15 20 25 300

2

4

6

8

n

Inte

gral

Figure 2.9: Absolute value of sinc function and its integral values for t = n sec.




Pr. 2.17 (a) We can either find the impulse response or show that the output is bounded for any input signalthat is bounded. To find the impulse response change the variable to σ = t− τ which gives

y(t) =−1

T

∫ 0

T

x(t− σ)dσ =

∫ T

0

1

Tx(t− σ)dσ

which is a convolution integral with h(t) = (1/T )[u(t) − u(t − T )]. This impulse response is absolutelyintegrable as ∫ ∞

0

|h(t)|dt =

∫ T

0

1

Tdt = 1

Likewise, if we assume that x(t) is bounded, i.e., there is a value M <∞ such that |x(t)| < M , then

|y(t)| ≤ 1

T

∫ t

t−T|x(τ)|dτ ≤ M

T

∫ t

t−Tdτ = M <∞

so system is BIBO stable.(b) The ramp is not bounded, there is no M such that |r(t)| < M < ∞. So when we compute the output ofthe stable system to the ramp we do not expect it to be bounded either. Indeed

y(t) =1

T

∫ t

t−Tτdτ =

t2 − (t− T )2

2T= t− T

2

increases as t increases, so it is unbound.




Pr. 2.18 (a) Assuming that the input is xs(0) when the switch is closed. If r is very small, the capacitor ischarged very fast, i.e., for 1/rC a large value, the voltage of the hold circuit from 0 ≤ t ≤ ∆ will be of theform

vc(t) ≈ xs(0)[1− e−t/rC ]u(t)

when the switch opens the resistor r and the constant source is disconnected, and the capacitor dischargesthrough the resistor R. The initial voltage is x ∗ (∆) which should be close to the sample value if Cr ischosen correctly. The voltage in the capacitor for ∆ ≤ t ≤ Ts is of the form

x∗(t) = x∗(∆)e−t/RC

and should go to zero as t approaches Ts, this can be accomplished by letting RC < Ts, so that the exponen-tial decays very fast.The signal from 0 to Ts is then a signal reaching a unit step of amplitude xs(0) by t = ∆ and it approximateszero for ∆ < t ≤ Ts.(b) If the input is a ramp, the sampled-and-hold signal at the output of the circuit is composed of pulsesdescribed in (a) each with amplitude approximating xs(nTs) and decaying.

xs(0)

Ts!

Ts 2Ts 3Ts 4Ts

txs(0)

xs(Ts)

xs(2Ts)

xs(3Ts)

xs(4Ts)x!(t)

x!(t)

Figure 2.10: Sample and hold outputs for a pulse and for a ramp signal.




Pr. 2.19

clear all; clf

% parts (a)--(c)

Ts=0.01;

t=0:0.01:100; N=length(t)

p=[20*ones(1,4000) -10*ones(1,2000) zeros(1,N-6000)];

P=1.1*abs(min(p))

x=(p+P).*cos(2*pi*t);

figure(6)

subplot(221)

plot(t,p); ylabel(’p(t)’)

subplot(222)

plot(t,x); ylabel(’x(t)’)

y=abs(x);

subplot(223)

plot(t,y); ylabel(’y(t)’)

h=exp(-0.8*t);

z=conv(h,y)*Ts; z=z*20/15;

subplot(224)

plot(t,z(1:length(t))-P); ylabel(’z(t)’); xlabel(’t’)

% part (d)

Ts=0.01;

t=0:0.01:100;N=length(t)

p=2*cos(0.2*pi*t);

P=abs(min(p))

x=(p+P).*cos(10*pi*t);

figure(7)

subplot(221)

plot(t,p+P); ylabel(’p(t)+P(t)’)

subplot(222)

plot(t,x); ylabel(’x(t)’)

y=abs(x);

subplot(223)

plot(t,y); ylabel(’y(t)’); xlabel(’t’)

h=exp(-0.8*t);

z=conv(h,y)*Ts;z=z-P+0.4;z=z*2/max(z);

subplot(224)

plot(t,z(1:length(t)));grid; ylabel(’z(t)’); xlabel(’t’)




0 20 40 60 80 100−20

−10

0

10

20

p(t)

0 20 40 60 80 100−40

−20

0

20

40

x(t)

0 20 40 60 80 100−20

−10

0

10

20

y(t)

0 20 40 60 80 100−20

−10

0

10

20

30

z(t)

t

0 20 40 60 80 1000

1

2

3

4

p(t)

+P

(t)

0 20 40 60 80 100−4

−2

0

2

4

x(t)

0 20 40 60 80 1000

1

2

3

4

y(t)

t0 20 40 60 80 100

−3

−2

−1

0

1

2

z(t)

t

Figure 2.11: Modulation and envelope detection for pulses and for sinusoid.




Pr. 2.20

clear all; clf

% part (a)

Ts=0.01;

t=0:Ts:10;

omega_c=2*pi; nu=10;

m=cos(t); %% message

im=sin(t); %% integral of message

y=cos(omega_c*t +2*pi*nu*im); %% FM1 with nu=10

y1=cos(omega_c*t +2*pi*nu*im/10); %% FM2 with nu/10

figure(9)

subplot(311)

plot(t,m); ylabel(’m(t)’)

subplot(312)


subplot(313)

plot(t,y1); ylabel(’y_1(t)’)

% part(b)

N=length(t)

for k=1:N,

if m(k)>=0,

m1(k)=1; %% message

y(k)=cos(omega_c*k*Ts +2*pi*nu*k*Ts/10); %% FM

else

m1(k)=-1;

y(k)=cos(omega_c*k*Ts -2*pi*nu*k*Ts/10);

end

end

figure(10)

subplot(211)

plot(t,m1); ; ylabel(’m_1(t)’)

axis([0 10 1.1*min(m1) 1.1*max(m1)])

subplot(212)


axis([0 10 1.1*min(y) 1.1*max(y)])




0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

m(t

)

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

y(t)

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

y 1(t)

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

m1(t

)

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1

y(t)

Figure 2.12: FM signals for ν = 10 and 1 (top), and for ν = 1 (bottom)



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