consider the lti system initially at rest and described by the difference equation

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Example 7.  consider LTI system initially at rest described by the difference equation

y[n]  – ay [n – 1] = x[n], |a| < 1

find the impulse response of this system.

Sol. the frequency response of the system is

H(ejo) = 1/1 – ae-jo

we can use the inverse fourier transform to get the impulse response

h[n] = anu [n]

Properties of the Discrete Time Fourier Transform

In this section, we use the following notation. let {x[n]} and {y[n]} be two signals, then their DTFT is denoted by x (ejo) and y(ejo). the notation

{x[n]} – x (ejo)

is used to say that left hand side is the signal x[n] whose DTFT is x (ejo) is given at right hand side.

  1. Periodicity of the DTFT

As noted earlier that the DTFT x(ejo) is a periodic function of 0 with period 2, this property is different from the contnuous time fourier transform of a signal.

  1. Linearity of the DTFT

if              {x[n]} – x (ejo)

and         {y[n]} – y (ejo)

then, a{x[n]} + b{y[n] – ax (ejo) + by (ejo)

this follows easily from the defining eq. (2).

  1. Conjugation of the signal

if     {x[n]} – x (ejo)

then,    {x* [n]} – x* (e – jo)

where* denotes the complex conjugate. we have DTFT of {x*[n]}

x* [n] e-jon = [x [n] ejon]*

= x[n] e-j(-0) n]

x * (e-jo)

  1. Time reversal

{x[-n]}  – x(e-jo)

the DTFT of the time reversal sequence is

x [-n] e-jon

let us change the index of summation as m = – n

x[m] ejom = x (e-jo)

  1. Symmetry Properties of Fourier transform

if  x[n] is real valued than

x(ejo ) = x* (e-jo)

The follows from property 3. if x[n] is real valued then x[n] = x* [n], so {x[n]} = {x* [n]} and hence

x{ejo) = x* (e-jo)

expressing x (ejo) in  real and imaginary parts we see that

xr (ejo) + jx1 (ejo) = xr (e-jo) – jx1 (e-jo)

which implies

xr (ejo) = xr (e-jo)

x1 (ejo) = – x1 (e-jo)

that is real pat of the fourier transform is an even function of o and imaginary part is an odd function of 0.

The magnitude spectrum given by

|x(ejo)| = x2r (ejo) + x21 (ejo)

= x2r  (e-jo) + x21 (e-jo) = |x (e-jo)|

Hence, magnitude spectrum of a real signal is an even function of o. the phase spectrum is given by

<x (ejo) = tan -1  x1 (ejo)/xr (ejo)

= tan-1  -x1 (e-jo)/xr (e -jo)

= – tan-1 x1 (e-jo)/xr (e-jo)

= – x (e-jo)

Thus, the phase spectrum is an odd function of 0. we denote the symmetric and anti-symmetric part of a function by

ev{{x[n]}) = 1/ {x[n]} + 1/2 {x* [-n]}

od ({x[n]}) = 1/2 {[x]} = 1/2 {x* [-n]}

ev (x(ejo)) = 1/2 x (ejo) + 1/2 x* (e-jo)

od (x(ejo)) = 1/2 x (ejo) – 1/2 x* (e-jo)

Then using properties (2) and (3) we see that

ev {x[n]} – re x(ejo)

d {x[n]} – jim x(ejo)

and using properties (2) and (4), we can see that

re ({x[n]}) – ev (x(ejo))

im ({x[n]}) – od (x(ejo))

  1. Time shifting and frequency shifting

{x[n – no]} – e-jono x(ejo)

{ejo 0n x[n]} – x (ej (0 – 00)

These can be proved very easily by direct substiution of x[n – no] in eq. (2) and x(ej (0 – 0)) in eq. (1).

  1. Differencing and summation

{x[n] – x[n – 1]} – [1 – e-jo) x (ejo)

this follows directlly from linearity property 2.

consider next the singal {y[n]} defined by

y[n] = x[m]

since, y[n] – y [n – 1] = x[n], we are lempted to conclude that the DTFT of {y[n]} is DTFT of {x[n]} divided by (1 – e-jo). this is not entirely true as it ignores the possibility of a DC or average term that can result from summation. the precise relationship is

x[m] 1/1 – e-jo x (ejo) + x(ejo) (0 + 2k)

we omit the proof of this property.

if we take {x[n]} = {[n]}, then we get

{u[n]} = [n] – 1/1 – e-jo + (0 + 2k)

  1. Time and frequency scaling

For continuos time signals, we know that the fourier transform of x(at) is given by 1/|a| x (a). however, if we define a signal {x[a]} we run into difficulty as the index must be can integer. thus, if a is an integer say a = k > 0. then we get signal {x[kn]}. this consists of taking kth sample of the original signal. thus, the DFTF of this signal looks similar to the fourier transform of a sampled signal. the result that resembles the continuous time signal is obtained if we define a signal {x(k)[n]} by

x(k) [n] = {x [n/k], if n is multiple of k

if n is not a multiple of k

for example, {x(2) [n]} is illustrated below.

The signal {x(k)[h]} is obtained by inserting (k – 1) zeros between successive value of signal {x[n]}.

x(k) (ejo)  x(k) [n] e-jon

= x[km] e-jokm

= x (ejko)

here, we can note the time frequency uncertainly. since, {x(k) [[n]} is expanded sequence, the fourier transform is compressed.

  1. Differentiation in frequency domain

x (ejo) = x[n] e-jon

differentiating both sides w.r.t, we obtain

d/d x (ejo) = – jnx [n] e-jon

multiplying both sides by j, we obtain

{nx[n]} – jd/d x (ejo)

  1. Parseval’s relation

|x[n]|2 = 1/2 |x(ejo)|2 d

we have,

|x[n]|2 = x[n] [1/2 x(ejo) ejon) d

interchanging summation and integration, we get

= 1/2 x* (ejo)  x[n] e-jon d

= 1/2  x* (ejo) x (ejo) d

= 1/2  |x(ejo)|2 d

  1. Convolution Property

This is the eigen function property of the complex exponential mentioned in the begining of the chapter. the fourier syntaxis eq. (1) for the x[n] can be interpreted as a representaion of {x[n]} in terms of linear combinations of complex exponential with amplitude proportional to x(ejo). each of these complex exponential is an eigen function of the LTI system and so the amplitude y(ejo) in the decomposition of {y[n]} will be x(ejo) H(ejo), where H(ejo) is the fourier transform of the impulse response. we prove this formally the output {y[n]} is given in terms of convo;ution sum so,

y (ejo) = y[n] e-jon

= (h[k] x [n – k]) e-jon

interchanging order of the summation

= h[h]  x[n – k] e-jon

let m = n – k then n = m + k and we get

= h[k]   x[m] e-jo(m + k)

= h[k] e-jok  x[m]e-jom

= h (ejo) x(ejo)

thus, if {y[n]} * {x[n]}

y (ejo) = h (ejo) x (ejo) ……………….(20)

Convolution in time domain becomes multiplication in the frequency domain. the fourier transform of the impulse response {h[n]} is known as frequency response of the system.

  1. The modulation of windowing property

let us find the DTFT of product of two sequences

{x[n]} {y[n]} = {x[n] y [n]} = {z[n]}

z(ejo) = x[n] y[n] e-jon

substituting for x[n] in terms of IDFT, we get

= (1/2  x (eja (ejan dx) y[n] e-j0n

interchanging order of integration and summation,

= 1/2  x (eja ) [ y[n] e-j(0 – a) n] da

= 1/2 x (eja) y (ej (0 – a) da

This looks like convolution of two functions, only the interval of integration is to x(ejo) and y (ejo) are periodic functions and eq. (21) is called periodic convolution. thus,

{x[x[n] y[n]} – 1/2 x(ejo) y[ejo)

where, denotes periodic convolution.

Example 6.  find the frequency response of systems characterized by linear constant with coefficient difference equation.

Sol. As we have seen earliear, constant coefficient linear difference eqyation with zero initial condition can be used to describe some linear time invariant systems.

the input-output {x[n]} and {y[n]} are reated by

ak y [n – k] = x[n – k]

we assume that fourier transforms of {x[n]}, {y[n]} and {h[n]} h(h) are the impulse response response of the system exist then convolution property implies that

h (ejo) = y(ejo)/x (ejo)

Taking fourier transform of both sides of eq. (22) and using linearity and time shifting properties of the fourier transform, we get

ak e-jok y(ejo) = bk e-jok x(ejo)

h(ejo) = y(ejo)/x(ejo)

bk e-jok /ak e-jok …………………..(23)

Thus, we see that the frequency response is the ratio of polynomials in the variable e-jo the numerator coefficients are the coefficients of x[n – k] in eq. (22) and denominator coefficients are the coefficients of y[n – k] in eq. (22). thus, we can write the frequency response by inspection.

Example 7.  consider LTI system initially at rest described by the difference equation

y[n]  – ay [n – 1] = x[n], |a| < 1

find the impulse response of this system.

Sol. the frequency response of the system is

H(ejo) = 1/1 – ae-jo

we can use the inverse fourier transform to get the impulse response

h[n] = anu [n]

Properties of the Discrete Time Fourier Transform

x[n] = 1/2  x(ejo) ejon d

x (ejo) = x[n] e-jon

Intro Exercise – 6

  1. The fourier transform of signal e-2t u(t – 3) is

(a) e-3(2 – jo)/2 – jo

(b) e-3(2 + jo)/2 + j

(c) e3(2 – jo)/2 – j

(d) e3(2 + jo)/2 + j

  1. The fourier transform of signal e-4|t| is

(a) 8/16 + 02

(b) -8/16 + 02

(c) 4/16 + 02

(d) -4/16 + 02

  1. The fourier transform of signal (t + 1) (1 – 1) is

(a) 2/1 + j

(b) 2/1 – j

(c) 2 cos

(d) none of these

  1. The fourier transform of signal te-1 u(t) is

(a) 1/1 + 02

(b) -1/1 + 02

(c) 1/(1 + j)2

(d) 1/(1 – j)2

  1. The fourier transform of signal am (t – m),

(a) <1, is

(b) a/1 + ae-jo

(c) 1/1 + ae-jo

(d) 1/1 – ae-jo

  1. The fourier transform of signa e-t + 2 u(t – 2 ) is

(a) -02/1 + j

(b) 02/1 + j

(c) e-j20/1 + 2

(d) ej2/1 + 2

  1. The fourier transform of signal (sin 2t) e-t u(t) is

(a) 1/2 [1/1 + j(0 – 2) – 1/1 + j (0 + 2)

(b) 1/2j [1/1 + j(0 – 2) – 1/1 + j(0 + 2)

(c) 1/2j [1/1 + j(0 + 2) – 1/1 + j(0 – 2)

(d) 1/2 [1/1 + j (0 + 2) – 1/1 + j (0 – 2]

  1. The fourier transform of signal te-3|t-1| is

(a) 6e-jo/9 + 02 – 12 je-jo/(9 + 02)2

(b) 12e-jo/(9 + 02)3

(c) 6e-j30/9 + 02 – 12j e-3jo/(9 + 02)2

(d) 12 j e-jo/(9 + 02)3

  1. The fourier transform of signal sgn(t) is

(a) -2/j

(b) 4/j

(c) 2/j

(d) 1/j + 1

  1. The fourier transform of signal u(t) is

(a) (0)

(b) 1/j

(c) (0) + 1/j

(d) none of these

  1. The fourier transform of signal 1/a2 + t2 is

(a) a ea|0|

(b) a e-a|0|

(c) 2/a e-a|0|

(d) 2/a ea|0|

  1. The fourier transform of signal shown below is

(a) 2 – 2e-2 sin 2 + 2 e-2 sin 2

(b) 2 + 2e-2 cos 2 – 2e-2 cos  2

(c) 2 – 2e-2 cos 2 + 2e-2 sin 2/1 + 02

(d) 2 + 2e-2 cos 2 – 2e-2 sin 2/1 + 02

  1. The fourier transform of signal shown below is

(a) 2sin 0 – 2/0

(b) 2 cos 0 – 2/j

(c) 2 j cos o

(d) 2 j sin

  1. The inverse fourier transform 2 (0) + (0 – 4) + (0 + 4) is

(a) 2 (1 – cos 4t)

(b) (1 – cos 4t)

(c) 1 + cos 4t

(d) 2 (1 + cos 4t)

  1. The inverse fourier transform of e-2|0| is

(a) 2/(4 + t2)

(b) 1/2 (4 + t2)

(c) 1/(4 + t2)

(d) none of these

  1. Consider the following signal

x[n] = {1, |n|<2 0, otherwise

the discrete time fourier transform above signal is

(a) sin 5/sin

(b) sin 4/sin

(c) sin 2.5 /sin 0.5

(d) none of these

  1. The DTFT of (3/4)n u[n – 4] is

(a) (3/4e-j)4/1 – 3/4 e-j

(b) (3/4 ej)4/1 – 3/4 ej

(c) (3/4 e-j)4/1 + 3/2 ej

(d) none of these

  1. The DTFT of u[n – 2] -u [n – 6] is

(a) e3j + e3j + e4j + e5j

(b) e-2j (1 – e3j)/1 – ej

(c) e-2j + e-3j + e-4j + e-5j

(d) e-2j (1 – e-3j)1 – e-j

  1. The DTFT of a|n|, |a| < 1 is

(a) 1 – a2/1 – 2a sin + a2

(b) 1 – a2/1 – 2a cos + a2

(c) 1 – a2/1 – 2ja sin + a2

(d) none of these

  1. The DTFT of (1/2)-n u[-n – 1] is

(a) ej/2 – e-j

(b) 2ej/2 – e-j

(c) ej/2 – ej

(d) 2ej/2 – ej

  1. The DTFT of 2 [(4 – 2n] is

(a) 2e-j2

(b) 2ej2

(c) 1

(d) none of these

Answers with Solutions

  1. (b)

x(jo) = x(t) e-jot dt = e-2te-jot dt

= e-3(2 + jo)/2 + j

  1. (a)

x(j) = e-4|r| e-jot dt

e-4|t| = {e-4t, t > 0   e4t , t > 0

= e4t  e-jot + e-4t e-jot dt

= 8/16 + 02

  1. (c)

x(t + t0) – ejot0 x(j)

x (j) = (t + 1) + (t – 1)] e-jot dt

= ejo + e-jo = 2 cos

  1. (c)

te-at u(t) – 1/(a + j)2

x(j) = te-t e-jot

= te-t(1 + jo) = 1/(1 + j)2

  1. (d)

x(j) = am (t – m) e-jot dt

= (ae-jo)m = 1/1 – ae-jo

  1. (c)

e-t u(t) – 1/1 + j

x1(t – 2)  – e-2jo x1 (j)   [using shifting property]

x(j) = e-j2o/1 + 2

  1. (b)

e-t u(t) – 1/1 + j

e-3|t| – 6/9 + 02

ej2t x1(t) – x1 {j(0 – 2)}

x(j) – 1/2j (1/1 + j(0 – 2) – 1/1 + j(0 + 2)

  1. (a)

e-3 |t| = {e-3t , t > 0  e3t , t > 0

e-3|t| – 6/9 + 02

x1 (t – 1) – e-jo x1(j)

tx2(t) – jd/d x2 (j)

x (j) = j d/d [e-jo 6/9 + 02] = 6e-jo/9 + 02 – 12 je-2jo/(9 + 02)

  1. (c)

sgn(t) = {1,  0 < t < 0   -1 – 0 < t < 0

x(0) = (-1)e-jot dt + (1) e-jot dt

= e-jot/j + e-jot/-j = 2/j

  1. (c)

1 – 2 (0)

1/2  (0)

sgn t – 2/j

1/2 sgn (t) – 1/j

1/2 + 1/2 sgn (t) = u(t) – (0) + 1/j

  1. (b)

e-a|t| – 2a/a2 + 02

e-a|t| = {e-at, t > 0  eat,  t < 0

using duality property,

2a/a2 + t2 – 2e-a|0| = 2e-a|0|

1/a2 + r2 – a e-a|0|

  1. (c)

x(j) = et e-jot dt +  e-t e-jot dt

= 1 – e-(1 – jo)2/1 -j + 1 – e-(1 + jo)2/1 + j

= 2 – 2e-2 cos 2 + 2e-2 sin 2/1 + 02

  1. (b)

x(j) = e-jot dt – e-jot dt = 2cos 0 – 2/j

  1. (c)

x(t) = 1/2  [2 (0) + (0 – 4) + (0 + 4)]ejot d

= 1 + 1/2 ej4t + 1/2 e-j4t = 1 + cos 4t

  1. (a)

e-2|0| = {e-20, 0 > 0  e20, 0 < 0

x(t) = 1/2 e-2|0|e-jot d

= 1/2  e2 ejot + 1/2  e-2 e-jot d = 2/(4 + t2)

  1. (c)

x(ej) = x[n] e-jn = e-jn

= ej2 + ej + 1 + e-j + e-j2

= e-j2 (1 + ej + ej2 + ej3 + ej4)

= e-j2 (1 – ej5)/1 – ej

= ej5/2 – ej5/2/e-j/2 – ej/2 = sin 2.5 /sin 0.5

  1. (a)

x(ej) = x[n]e-jn

= (3/4) nejn = (3/4 e-j)n = (3/4 e-j)/1 – 3/4 e-j

  1. (c)

x[n] = u[n – 2] – u[n – 6]

= [n – 2] + [n – 3] + [n – 4] + [n – 6]

x(ej) = x[n] ejn

= e-2j + e-3j + e-4j + e-5j

  1. (b)

x(ej) = a-n ejn + ane-jn

= (aej)-n + (ae-j)n

= aej/1 – aej + 1/1 – ae-j = (1 – a2)/(1 + a2 – 2a cos )

  1. (c)

x(ej) = (1/2)-n u[-n – 1] ejn

= (1/2)-n ejn = (e-j/2)-n

= (ej/2)n

= 1/2 ej/1 – 1/2 ej = ej/2 – ej

  1. (a)

x(ej) = 2[4 – 2n] e-jn = 2e-j2