6.6 (b) the impulse response h1[n] is as shown in figure s6.6,as was increase ,it is clear that the significant central lobe of h1[n] becomes more concentrated around the origin. consequently. h[n]=h1[n](-1)^n also becomes more concentrated about the origin.
6.7 the frequency response magnitude |H(jw)| is as shown in figure s6.7.the frequency response of the bandpass filter G(jw) will be given by
This is as shown in figure s6.7
G（j）
-6000 -4000 -2000 0 2000 4000 6000
Figure S6.7
(a)from the figure ,it is obvious that the passband edges are at 2000∏rad/sec and 6000∏rad/sec. this translates to 1000HZ and 3000Hz,respectively.
(b)(b)from the figure ,it is obvious that the stopband edges are at 1600∏ rad/sec.this translates to 800Hz and 3200 Hz, respectively.
6.8 taking the Fourier transform of both sides of the first difference equation and simplifying, we obtain the frequency response H(e^jw)of the first filter.
Taking the Fourier transform of both sides of the second difference equation and simplifying ,we obtain the frequency response H1(e^jw) of the second filter.
This may also be written as
Therefore .the frequency response of the second filter is obtained bu shifting the frequency response of the first filter by ∏.although the first fitter has its passband between-wp and wp. Therefore, the second filter will have its passband between ∏-wp and ∏+wp.
6.9 taking the Fourier transform of the given differential equation and simplifying .we obtain the frequency of the LTI system to be
Taking the inverse Fourier transform, we obtain the impulse response to be
Using the result derived in section 6.5.1,we have the step response of the system
The final value of the step response is
We also have
Substituting s(t0)=(2/5)[1-1/e^2],in the above equation ,we obtain t0=2/5 sec
(a)we may rewrite H1(jw)to be
we may then treat each of the two factors as individual first order systems and draw their bode magnitude plots .the final bode will then be asum of these two bode plots .this is shown in the figures6.10
mathematically. the straight-line approximation of the bode magnitude plot is
Figure S6.10
(b) Using a similar approach as in part (a),we obtain the Bode plot to be as shown in
Figure S6.10.
Mathematically, the straight-line approximation of the BODE magnitude plot is
6.10. (a) We may rewrite the given frequency response (j) as
.
We may then use an approach similar to the one used in Example 6.5 and in Problem
6.11 to obtain the Bode magnitude plot(with straight line approximations) shown in
Figure S6.11.
Mathematically, the straight-line approximation of the Bode magnitude plot is
(b)We may rewrite the frequency response ( j) as
=.
Again using an approach similar to the one used in Example 6.5,we may draw the
Bode magnitude plot by treating the first and second order factors separately. This
Givens us a Bode magnitude plot (using straight line) approximations as shown below:
Mathematically , the straight-line approximation of the Bode magnitude plot is
6.12. Using the Bode magnitude plot, specified in Figure P6.12(a). we may obtain an expression
For (j). The figure shows that (j) has the break frequencies=1, =8,And =40. The frequency response rises as 20dB/decade after. At,this rise is canceled by a -20 dB/decade contribution. Finally, at ,an additional -20 dB/decade. Contribution results in the subsequent decay at the rate of -20 dB/decade, therefore, we may conclude that
=.
We now need to find A. Note that when =0, 20=2.Therefore. =0.05. From eq . (S6.12-1),we know that
=.
Therefore, A =0. This gives us
=.
Using a similar approach on Figure P6.12(b), we obtain
H(j)=.
Since the overall system (with frequency response H(j)) is constructed by cascading
Systems with frequency responses and ,
H(j)=.
Using the previously obtained expressions for H(j) and ,
= H(j)/=.
6.13. Using an approach similar to the one used in the previous problem, we obtain
H(j)= .
(a)Let us assume that we desire to construct this system by cascading two systems with
frequency responses and , respectively. We require that
H(j)=.
We see that = and =
And
= and =
are both valid combinations.
(b) Let us assume that we desire to construct this system by connecting two systems with
frequency responses and in parallel. We require that
H(j) = +
Using partial fraction expansion on H(j) , we obtain
H(j) = -
From the above expression it is clear that we can define and in only one way
6.14. Using an approach similar to the one used in Problem 6.12 ,we have
H(j)=.
The inverse to this system has a frequency response
= =.
6.15. We will use the results from Section 6.5 in this problem.
(a) We may write the frequency response of the system described by the given differential
Equation as
=.
This may be rewritten as
=.
From this we obtain the damping ratio to be =1.Therefore , the system is critically damped
(b)We may write the frequency response of the system described by the given differential equation as
=.
This may be rewritten as
=.
From this we obtain the damping ratio to be =2/5. Therefore , the system is under-damped.
(c)We may write the frequency response of the system described by the given differential equation as
=.
This may be rewritten as
=.
From this we obtain the damping ratio to be =10. Therefore , the system is under-damped.
(d)We may write the frequency response of the system described by the given differential equation as
=.
The terms in the numerator do not affect the ringing behavior of the impulse response of this system .Therefore , we need to only consider the denominator in order to determine if the system.
is critically damped, under-damped ,or over-damped .We see that this frequency response has the same denominator as the one obtained in part (b).Therefore . this system is still under-damped.
6.16. The system of interest will have a difference equation of the form
y[n] –ay[n-1] =b x [n].
Making slight modifications to the results obtained in Section 6.6.1,we determine the step response of this system to be
.
The final value of the step response will be b/(1-a). The step response exhibits oscillatory behavior only if <1.Using this fact, we may easily show that the maximum overshoot in the step response occurs when n=0. Therefore , the maximum value of the step response is
=b.
Since we are given that the maximum, overshoot is 1.5 times the final value, we have
1.5=b a=
Also ,since we are given that the final value us 1,
=1 b=
Therefore, the difference equation relating the input and output will be
y[n]+ y[n-1]= x[n].
6.17. We will use the results derived in Section 6.6.2 to solve this problem.
(a)Comparing the given difference equation with eq. (6.56),we obtain
=, and =-1.
Therefore, =,and the system has an oscillatory step response.
(b) Comparing the given difference equation with eq. (6.56),we obtain
=, and =-1.
Therefore, =0,and the system has non- oscillatory step response.
6.18. Let us first find the differential equation governing the input and output of this circuit.
Current through capacitor =C.
Voltage across resistor = RC.
Total input voltage =Voltage across resistor + Voltage across capacitor
Therefore ,
x(t)= RC +y(t).
The frequency response of this circuit is therefore
H(j)=.
Since this is a first order system , the step response has to be non oscillatory.
6.19. Let us first find the differential equation governing the input and output of this circuit .
Current through resistor and inductor = Current through capacitor = C.
Voltage across resistor = RC.
Voltage across inductor = LC.
Total input voltage = Voltage across inductor + Voltage across resistor + Voltage across capacitor
Therefore ,
x(t)= LC+ RC +y(t).
The frequency response of this circuit is therefore
H(j)=.
We may rewrite this to be
H(j)=.
Therefore, the damping constant =. In order for the step response to have no
oscillations, we must have 1 .Therefore, we require
R2.
6.20. Let us call the given impulse response h[n]. It is easily observed that the signal [n]=h[n+2] is real and even . Therefore ,(using properties of the Fourier transform ) we know that the Fourier transform of [n] is real and even . Therefore has zero phase, we have
=.
Therefore , the group delay is
=2.
6.21. Note that in all parts of this problem Y(j)= H(j)X(j)=X(j). Therefore ,y(t)= .
(a) Here, x(t)=.Therefore, y(t)=.This part could also have been solved by nothing that complex exponentials are Eigen functions of LTI systems. Then when x(t)=,y(t) should be y(t)=H(j1) =-2j.
(b)Here ,x(t)=sin()u(t).Then , = (t)u(t)+ (t) (t)= (t)u(t).
Therefore ,y(t)= 2=2 (t)u(t).
(c) Here , Y(j)=X(j)H(j)=2/(6+j) .Taking the inverse Fourier transform we obtain y(t)=.
(d)Here, x(jw)=1/(2+jw). From this we obtain x(t)= u(t). Therefore, y(t)= -2dx(t)/dt=4u(t) - 2 (t).
6.22 Note that H(jw)=
(a)Since x(t)=cos(2t+),X(jw)= (w-2)+ (w+2). This is zero outside the region -3< w<3.Thus, Y(jw)=H(jw)X(jw)=(jw/3)X(jw). This implies that y(t)=(1/3)dx(t)/dt=(-2/3)sin(2t+).
(b)Since x(t)= cos(4t+),X(jw)= (w-4)+ (w+4). Therefore, the nonzero portions of X(jw) lie outside the range -3< w<3. This implies that Y(jw)=H(jw)X(jw)=0.Therefore,y(t)=0.
(c)The Fourier series coefficients of the signal x(t) are given by
= x(t)
Where =1 and =2/=2, Also,
X(jw)=2
The only impulses of X(jw) which lie in the region -3< w<3 are at w=0,2,and 2.Defining the signal (t)= =1/, ==-1/(4j).Putting these into the expression for (t) we obtain (t)=(1/)+(1/2)sin(2t). Finally, y(t)=(1/3)d (t)/dt=(1/3)cos(2t).
6.23. (a) From the given information, we have
(jw)=
Using Table 4.2, we get
(t)=.
(b)Here,
= (jw)
Using Table 4.1, we get
= (t+T)
Therefore,
(t)=
(c)Let us consider a frequency response (jw) given by
(jw)=
Clearly,
(jw)= [ (jw)*W(jw)],
Where
W(jw)=j2 (w-)-j2 (w-)
Therefore , from Table 4.1
(t)= (t)w(t)=[][-2sin(t/2)].
6.24. If (w)=, where is a constant, then
H(jw)=- w+
Where is another constant.
(a)Note that if h(t) is real, the phase of the Fourier transform H(jw) has to be an odd function. Therefore, the value of in eq. (S6.24-1) will be zero.
Also, let us define (jw)=|H(jw)|. Then
(t)=
(i)Here =5. Hence, H(jw)=-5w. Then
H(jw)=|H(jw)| = (jw)
Therefore,
H(t)= (t-5)=
(ii)Here =5/2. Hence, H(jw)=-(5/2)w. Then,
H(jw)=|H(jw)| = (jw)
Therefore,
h(t)= (t-5/2)=
(iii)Here =-5/2. Hence, H(jw)=(5/2)w. Then,
H(jw)=|H(jw)| = (jw)
Therefore,
h(t)= (t+5/2)=
(b)If h(t) is not specified to be real, then H(jw) does not have to be an odd function. Therefore, the value of in eq. (S6.24-1) does not have to be zero. Given only |H(jw)| and (w), cannot be determined uniquely. Therefore, h(t) cannot be determined uniquely.
6.25 (a) We may write (jw) as
(jw)=
Therefore,
(jw)=
And
Since, is not a constant for w. Therefore, the frequency respons has nonlinear phase.
(b)In the case, is the frequency response of a system which is a cascade combination of two system, each of which has a frequency response (jw). Therefore.
And
Since, is not a constant for all w. Therefore, the frequency response has nonlinear phase.
(c)IN this case, is again the frequency response of a system which is a cascade combination of two systems. The first system has a frequency response . While the second system has a frequency response . Therefore,
And
Since, is not a constant for all w . Therefore, the frequency response has nonlinear phase.
6.26. (a) Note that =1-, where is
=
Therefore,
h(t)= (t)-
From Table 4.2 .we have
Therefore,
h(t)= (t)-
(b)A sketch of is Figure S6.26. Clearly, as increase. h(t) becomes more concentrated about the origin.
(c) Note that the step response is given by
S(t)=h(t)*u(t)=u(t)-u(t)*
Also, note that is the impulse response of an ideal lowpass filter. If = u(t)* denotes the step response of the lowpass filter, we know from Figure 6.14 that =0 and =1. Therefore,
S(0+)=u(0+)-=1-(1/2)=1/2
And
S()=u()-
6.27. (a) Taking the Fourier transform of both sides of the given differential equation, we obtain
H(jw)=
The Bode plot is as shown in Figure S6.27
(b)From the expression for H(jw) we obtain
Therefore,
(c)Since x(t)=,
X(jw)=
Therefore,
Y(jw)=X(jw)H(jw)=
(d)Taking the inverse Fourier transform of the partial fraction expansion of Y(jw), we obtain
y(t)=
(e)(i) Here,
Y(jw)=
Taking the inverse Fourier transform of the partial expansion of Y(jw), we obtain
y(t)=
(ii) Here,
Y(jw)=
Taking the inverse Fourier transform of Y(jw), we obtain
y(t)=
(iii)Here,
Y(jw)=
Taking the inverse Fourier transform of the partial expansion of Y(jw), we obtain
y(t)= +
6.28. (a) The Bode plots are as shown below
(b) We may write the frequency response of (iv) as
H(jw)=
Therefore.
h(t)=
and
s(t)=h(t)*u(t)=
Both h(t) and s(t) are as shown in Figure S6.28.
We may write the frequency response of (vi) as
H(jw)=
Therefore,
h(t)=
and
s(t)=h(t)*u(t)=
Both h(t) and s(t) are as shown in Figure S6.28
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