Chapter 2

Review of Fundamentals of DigitalSignal Processing

2.1 (a) This system is not linear (the constant term makes it non linear) but is shift-invariant

(b) This system is linear but not shift-invariant (since the modulation term is not shift-invariant)

(c) This system is not linear (because of the cubic power) but is shift-invariant

(d) This system is linear and shift invariant (in fact the digital system is the convolution of theinput with a rectangular window of length N samples.

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2.2 (a) The system y[n] = x[n]+2x[n+1]+3 is not linear, as seen by the following counter example.Consider inputs x1[n] and x2[n] with outputs:

y1[n] = T [x1[n]] = x1[n] + 2x1[n+ 1] + 3

y2[n] = T [x2[n]] = x2[n] + 2x2[n+ 1] + 3

If we apply the system to the input x3[n] = ax1[n] + bx2[n] we get an output, y3[n] of theform:

y3[n] = T [ax1[n] + bx2[n]] = [ax1[n] + bx2[n]] + [2ax1[n+ 1] + 2bx2[n+ 1]] + 3

which is not equal to the linear sum ay1[n] + by2[n] thereby showing that the system is notlinear.

(b) The system y[n] = x[n]+2x[n+1]+3 is time-invariant (shift-invariant) as seen by consideringthe responses to x[n] and to x[n− n0], i.e.,

y[n] = T [x[n]] = x[n] + 2x[n+ 1] + 3

y[n− no] = T (x[n− n0]) = x[n− n0] + 2x[n− n0 + 1] + 3

(c) The system y[n] = x[n] + 2x[n+ 1] + 3 is not causal since the output at time n depends onthe output at a future time n+ 1.

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2.3 (a) The input sequence an is an eigen-function of LTI systems. Therefore if this system is LTI,the output must be of the form A· input or y[n] = Aan where A is a complex constant.Since bn 6= Aan for any complex constant A, the system cannot be LTI.

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4 CHAPTER 2. REVIEW OF FUNDAMENTALS OF DIGITAL SIGNAL PROCESSING

(b) The system is not LTI.

(c) In this case, the input excites the system at all frequencies (since it exists only for n ≥ 0).Therefore the system transfer function describes how the system transforms all inputs. Thusthis system could be LTI and there is only one LTI system with the given transfer function,i.e.,

H(z) =1− az−1

1− bz−1, |z| > b

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2.4 (a) x[n] can be written in the form:x[n] = anu[n− n0]

where

u[n− n0] =

{1 n ≥ n0

0 n < n0.

We can now solve for X(z) by the following steps:

x[n] = an−n0+n0u[n− n0]

= an0an−n0u[n− n0]

X(z) = an0

∞∑n=n0

a(n−n0)z−n

We can now make a change of variables to the form: n′ = n− n0 giving:

X(z) = an0

∞∑n′=0

an′z−n

′−n0

= an0z−n0

∞∑n′=0

an′z−n

′

=an0z−n0

1− az−1; |az−1| < 1

where we have used the relation:

∞∑n=0

rn =1

1− r; |r| < 1

(b) X(ejω) = X(z) evaluated on the unit circle (i.e., for z = ejω). Thus we get:

X(ejω) =an0e−jωn0

1− ae−jω; |ae−jω| < 1,

or equivalently, |a| < 1. The Fourier transform exists when the z-transform converges in aregion including the unit circle; in this case when |a| < 1.

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2.5 (a) x[n] can be written in the form:

x[n] = x1[n] + x2[n]

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where

x1[n] = u[n]

and

x2[n] = −u[n−N ] = −x1[n−N ]

Using this form for x[n], we can solve for the convolution output as the output due to u[n]for the region 0 ≤ n ≤ N−1, and as the output due to u[n]−u[n−N ] for the region N ≤ n.

We call the output of the convolution of x1[n] with h[n] as y1[n] and we solve for its valuein the region 0 ≤ n ≤ N − 1 using the convolution formula:

y1[n] = u1[n] ∗ h[n]

=∞∑

m=−∞x1[m]h[n−m]

=

n∑m=0

an−m(1)

= ann∑

m=0

a−m

= an1− a−n−1

1− a−1

=1− an+1

1− a0 ≤ n ≤ N − 1

We trivially solve for y2[n] = −y1[n−N ] as

y2[n] = − (1− an+1)

(1− a)N ≤ n

giving, for y[n], the value (for the region N ≤ n)

y[n] = y1[n] + y2[n] =1− an+1

1− a− 1− an−N+1

1− a

or, equivalently,

y[n] = an(a−N+1 − a)

(1− a)N ≤ n

(b) Using z−transforms we solve for Y (z) again as a sum in the form:

Y (z) = X(z) ·H(z)

= X1(z) ·H(z) +X2(z) ·H(z)

where X1(z) and X2(z) are the z−transforms, respectively, of x1[n] and x2[n] of part (a) ofthis problem. The resulting set of z− transforms is:

X1(z) =1

1− z−1

X2(z) = − z−N

1− z−1

H(z) =1

1− az−1

6 CHAPTER 2. REVIEW OF FUNDAMENTALS OF DIGITAL SIGNAL PROCESSING

We can now solve for Y1(z) = X1(z) ·H(z) giving the form:

Y1(z) =1

(1− z−1)(1− az−1)

Using the method of partial fraction expansion we factor Y1(z) into

Y1(z) =A

1− z−1+

B

1− az−1

We can now solve for A and B using the fact that the combined numerator is 1, giving:

A =1

1− a, B =

a

1− aNow we can invert the partial fraction expansion, giving

y1[n] =1

1− au[n]− a

1− aanu[n] =

1− an+1

1− au[n]

which is valid for all n but applies to the region 0 ≤ n ≤ N − 1. Similarly we can triviallysolve for y2[n] = −y1[n−N ]u[n−N ] again giving the same total result for the region N ≤ n.

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2.6 (1) The z−transform of the exponential window is of the form:

W1(z) =N−1∑n=0

anz−n =(1− aNz−N )

(1− az−1)

Note that zeros occur at zk = aej2πk/N , k = 1, 2, . . . , N − 1.

The Fourier transform for the exponential window is just:

W1(ejω) =(1− aNe−jωN )

1− ae−jω)

(2) The rectangular window is a special case of the exponential window with a = 1. The zerosare now all on the unit circle at zk = ej2πk/N , k = 1, 2, . . . , N − 1. The Fourier transform ofthe rectangular window is of the form:

W2(ejω) =(1− ejωN )

(1− ejω)= e−jω(N−1)/2 sin(ωN/2)

sin(ω/2)

(3) The Hamming window can be expressed in terms of shifted sums of rectangular windows, i.e.,

w3[n] = 0.54w2[n]− 0.23w2[n]ej2πn/(N−1) − 0.23w2[n]e−j2πn/(N−1)

The Fourier tranform of the Hamming window is thus of the form:

W3(ejω) = 0.54W2(ejω)

−0.23W2(ej(ω−2π/(N−1))− 0.23W2(ej(ω+2π/(N−1))

which can be put into the form

W3(ejω) = e−jω(N−1)/2[−0.54sin(ωN/2)

sin(ω/2)

+0.23sin[(ω − 2π/(N − 1))(N/2)]

sin[(ω − 2π/(N − 1))(1/2)]

+0.23sin[(ω + 2π/(N − 1))(N/2)]

sin[(ω + 2π/(N − 1))(1/2)]]

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Figure P2.6.1: Magnitude response of rectangular window

Figure P2.6.2: Magnitude response of exponential window

Figure P2.6.3: Magnitude response of Hamming window

A plot of |W2(ejω)| is shown in Figure P2.6.1. Notice that W1(z) has zeros on the unit circle;whereas |W1(z) has zeros on a circle of radius a as seen in Figure P2.6.2. Finally the Hammingwindow magnitude response is shown in Figure P2.6.3. We see how the side lobes tend to canceland that the main lobe is about twice the width of the rectangular window response.

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2.7 (a) A plot of an N = 9-point triangular window is shown in Figure P2.7.1.

Figure P2.7.1: 9-point triangular window

(b) An N−point triangular window can be created by convolving an (N+1)/2-point rectangularwindow with itself, i.e.,

wT [n] = wR[n] ∗ wR[n].

8 CHAPTER 2. REVIEW OF FUNDAMENTALS OF DIGITAL SIGNAL PROCESSING

Since convolution in time is equivalent to multiplication in frequency,we have:

WT (ejω) =[WR(ejω)

]2WR(ejω) =

1− e−jω(N+1)/2

1− e−jω= e−jω(N−1)/4 sin[ω(N + 1)/4]

sin(ω/2)

WT (ejω) = e−jω(N−1)/2

[sin[ω(N + 1)/4]

sin(ω/2)

]2

(c) Plots of the time and frequency (log magnitude) responses of a 101-point triangular windoware shown in Figure P2.7.2.

Figure P2.7.2: Time and frequency responses of 101-point triangular window

(d) The rectangular, Hamming and triangular windows compare as follows:

1. rectangular window: cutoff frequency= 1/N in normalized frequency units and is Fs/Nin analog frequency units, with sidelobe rejection ≥ 14 dB

2. Hamming window: cutoff frequency= 2/N in normalized frequency units and is 2Fs/Nin analog frequency units, with sidelobe rejection ≥ 44 dB

3. triangular window: cutoff frequency= 2/N in normalized frequency units and is 2Fs/Nin analog frequency units, with sidelobe rejection ≥ 28 dB

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2.8 (a) The impulse response of the ideal lowpass filter is obtained as:

h[n] =1

2π

∫ π

−πH(ejω)ejωndω =

1

2π

∫ ωc

−ωcejωndω

=1

2π

[ejωn

jn

]ωc−ωc

=1

2πjn[ejωcn − e−jωcn]

=sin(ωcn)

πn

(b) if ωc = π/4 then h[n] =1

4

sin(πn/4)

πn/4and a plot of h[n] is as shown in Figure P2.8.1.

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Figure P2.8.1: Impulse response of ideal lowpass filter.

Figure P2.8.2: Parallel combination of ideal filters.

(c) One apporach to obtaining the desired impulse response is to view H(ejω) as a parallelcombination of lowpass, highpass, and zero phase filters, as shown in Figure P2.8.2.

We can express HHP (ejω) in terms of a lowpass filter with cutoff frequency π−ωb with thepassband shifted by π. From Part (a) we have:

HLP (ejω)←→ sin(ωan)

πn

where ωa is the cutoff, giving:

HHP (ejω) = HLP (ej(ω−π))←→ ejπnsin[(π − ωb)n]

πn

where π − ωb is the cutoff frequency.

The allpass has the property:

HAP (ejω) = 1←→ δ[n]

Putting it all together we get:

h[n] = δ[n]− sin(ωan)

πn− ejπn sin[(π − ωb)n]

πn

= δ[n]− sin(ωan)

πn− (−1)n

sin[(π − ωb)n]

πn

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10 CHAPTER 2. REVIEW OF FUNDAMENTALS OF DIGITAL SIGNAL PROCESSING

(d) When ωa = π/4 and ωb = 3π/4, we can express h[n] for the bandpass filter as:

h[n] = δ[n]− 0.25 sin(πn/4)

πn/4− (−1)n

sin[(π − 3π/4)n]

πn

= δ[n]− 0.25 sin(πn/4)

πn/4− (−1)n

0.25 sin(πn/4)

πn/4

= δ[n]− 0.25 sin(πn/4)

πn/4[1 + (−1)n]

A plot of h[n] for the bandpass filter is shown in Figure P2.8.3.

Figure P2.8.3: Impulse response of ideal bandpass filter.

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2.9 (a) The magnitude response of an ideal differentiator is:

|H(ejω)| = |ω|

and the phase response is:

argH(ejω) =

{−ωτ + π/2 ω > 0

−ωτ − π/2 ω < 0

A plot of the magnitude and phase is given in Figure P2.9.1.

Figure P2.9.1: Magnitude and phase responses of ideal differentiator.

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