**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(e^{jo}) = 1/1 – ae^{-jo}

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

h[n] = a^{n}u [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 (e^{jo}) and y(e^{jo}). the notation

{x[n]} – x (e^{jo})

is used to say that left hand side is the signal x[n] whose DTFT is x (e^{jo}) is given at right hand side.

**Periodicity of the DTFT**

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

**Linearity of the DTFT**

if {x[n]} – x (e^{jo})

and {y[n]} – y (e^{jo})

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

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

**Conjugation of the signal**

if {x[n]} – x (e^{jo})

then, {x* [n]} – x* (e^{ – jo})

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

x* [n] e^{-jon} = [x [n] e^{jon}]*

= x[n] e^{-j(-0) n}]

x * (e^{-jo})

**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] e^{jom} = x (e^{-jo})

**Symmetry Properties of Fourier transform**

if x[n] is real valued than

x(e^{jo} ) = 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{e^{jo}) = x* (e^{-jo})

expressing x (e^{jo}) in real and imaginary parts we see that

x_{r} (e^{jo}) + jx_{1} (e^{jo}) = x_{r} (e^{-jo}) – jx_{1} (e^{-jo})

which implies

x_{r} (e^{jo}) = x_{r} (e^{-jo})

x_{1} (e^{jo}) = – x_{1} (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(e^{jo})| = x^{2}_{r} (e^{jo}) + x^{2}_{1} (e^{jo})

= x^{2}_{r } (e^{-jo}) + x^{2}_{1} (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 (e^{jo}) = tan ^{-1} x_{1} (e^{jo})/x_{r} (e^{jo})

= tan^{-1} -x_{1} (e^{-jo})/x_{r} (e ^{-jo})

= – tan^{-1} x_{1} (e^{-jo})/x_{r} (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(e^{jo})) = 1/2 x (e^{jo}) + 1/2 x* (e^{-jo})

od (x(e^{jo})) = 1/2 x (e^{jo}) – 1/2 x* (e^{-jo})

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

ev {x[n]} – re x(e^{jo})

d {x[n]} – jim x(e^{jo})

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

re ({x[n]}) – ev (x(e^{jo}))

im ({x[n]}) – od (x(e^{jo}))

**Time shifting and frequency shifting**

{x[n – n_{o}]} – e^{-jono} x(e^{jo})

{e^{jo 0n} x[n]} – x (e^{j (0 – 0}_{0})

These can be proved very easily by direct substiution of x[n – n_{o}] in eq. (2) and x(e^{j (0 – 0)}) in eq. (1).

**Differencing and summation**

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

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 (e^{jo}) + x(e^{jo}) (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)

**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)} (e^{jo}) x_{(k)} [n] e^{-jon}

= x[km] e^{-jokm}

= x (e^{jko})

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

**Differentiation in frequency domain**

x (e^{jo}) = x[n] e^{-jon}

differentiating both sides w.r.t, we obtain

d/d x (e^{jo}) = – jnx [n] e^{-jon}

multiplying both sides by j, we obtain

{nx[n]} – jd/d x (e^{jo})

**Parseval’s relation**

|x[n]|^{2} = 1/2 |x(e^{jo})|^{2} d

we have,

|x[n]|^{2} = x[n] [1/2 x(e^{jo}) e^{jon}) d

interchanging summation and integration, we get

= 1/2 x* (e^{jo}) x[n] e^{-jon} d

= 1/2 x* (e^{jo}) x (e^{jo}) d

= 1/2 |x(e^{jo})|^{2} d

**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(e^{jo}). each of these complex exponential is an eigen function of the LTI system and so the amplitude y(e^{jo}) in the decomposition of {y[n]} will be x(e^{jo}) H(e^{jo}), where H(e^{jo}) 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 (e^{jo}) = 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 (e^{jo}) x(e^{jo})

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

y (e^{jo}) = h (e^{jo}) x (e^{jo}) ……………….(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.

**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(e^{jo}) = x[n] y[n] e^{-jon}

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

= (1/2 x (e^{ja} (e^{jan} dx) y[n] e^{-j0n}

interchanging order of integration and summation,

= 1/2 x (e^{ja} ) [ y[n] e^{-j(0 – a) n}] da

= 1/2 x (e^{ja}) y (e^{j (0 – a}) da

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

{x[x[n] y[n]} – 1/2 x(e^{jo}) y[e^{jo})

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

a_{k} 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 (e^{jo}) = y(e^{jo})/x (e^{jo})

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

a_{k} e^{-jok} y(e^{jo}) = b_{k} e^{-jok} x(e^{jo})

h(e^{jo}) = y(e^{jo})/x(e^{jo})

b_{k} e^{-jok} /a_{k} 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(e^{jo}) = 1/1 – ae^{-jo}

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

h[n] = a^{n}u [n]

**Properties of the Discrete Time Fourier Transform**

x[n] = 1/2 x(e^{jo}) e^{jon} d

x (e^{jo}) = x[n] e^{-jon}

**Intro Exercise – 6**

- 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) e^{3(2 – jo})/2 – j

(d) e^{3(2 + jo})/2 + j

- The fourier transform of signal e
^{-4|t|}is

(a) 8/16 + 0^{2}

(b) -8/16 + 0^{2}

(c) 4/16 + 0^{2}

(d) -4/16 + 0^{2}

- 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

- The fourier transform of signal te
^{-1}u(t) is

(a) 1/1 + 0^{2}

(b) -1/1 + 0^{2}

(c) 1/(1 + j)^{2}

(d) 1/(1 – j)^{2}

- The fourier transform of signal a
^{m}(t – m),

(a) <1, is

(b) a/1 + ae^{-jo}

(c) 1/1 + ae^{-jo}

(d) 1/1 – ae^{-jo}

- The fourier transform of signa e
^{-t + 2 }u(t – 2 ) is

(a) -0^{2}/1 + j

(b) 0^{2}/1 + j

(c) e^{-j20}/1 + 2

(d) e^{j2}/1 + 2

- 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]

- The fourier transform of signal te
^{-3|t-1|}is

(a) 6e^{-jo}/9 + 0^{2} – 12 je^{-jo}/(9 + 0^{2})^{2}

(b) 12e^{-jo}/(9 + 0^{2})^{3}

(c) 6e^{-j30}/9 + 0^{2} – 12j e^{-3jo}/(9 + 0^{2})^{2}

(d) 12 j e^{-jo}/(9 + 0^{2})^{3}

- The fourier transform of signal sgn(t) is

(a) -2/j

(b) 4/j

(c) 2/j

(d) 1/j + 1

- The fourier transform of signal u(t) is

(a) (0)

(b) 1/j

(c) (0) + 1/j

(d) none of these

- The fourier transform of signal 1/a
^{2}+ t^{2}is

(a) a e^{a|0|}

(b) a e^{-a|0|}

(c) 2/a e^{-a|0|}

(d) 2/a e^{a|0|}

- 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 + 0^{2}

(d) 2 + 2e^{-2} cos 2 – 2e^{-2} sin 2/1 + 0^{2}

- 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

- 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)

- The inverse fourier transform of e
^{-2|0|}is

(a) 2/(4 + t^{2})

(b) 1/2 (4 + t^{2})

(c) 1/(4 + t^{2})

(d) none of these

- 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

- The DTFT of (3/4)
^{n}u[n – 4] is

(a) (3/4e^{-j})^{4}/1 – 3/4 e^{-j}

(b) (3/4 e^{j})^{4}/1 – 3/4 e^{j}

(c) (3/4 e^{-j})^{4}/1 + 3/2 e^{j}

(d) none of these

- The DTFT of u[n – 2] -u [n – 6] is

(a) e^{3j} + e^{3j} + e^{4j} + e^{5j}

(b) e^{-2j} (1 – e^{3j})/1 – e^{j}

(c) e^{-2j} + e^{-3j} + e^{-4j} + e^{-5j}

(d) e^{-2j} (1 – e^{-3j})1 – e^{-j}

- The DTFT of a
^{|n|}, |a| < 1 is

(a) 1 – a^{2}/1 – 2a sin + a^{2}

(b) 1 – a^{2}/1 – 2a cos + a^{2}

(c) 1 – a^{2}/1 – 2ja sin + a^{2}

(d) none of these

- The DTFT of (1/2)
^{-n}u[-n – 1] is

(a) e^{j}/2 – e^{-j}

(b) 2e^{j}/2 – e^{-j}

(c) e^{j}/2 – e^{j}

(d) 2e^{j}/2 – e^{j}

- The DTFT of 2 [(4 – 2n] is

(a) 2e^{-j2}

(b) 2e^{j2}

(c) 1

(d) none of these

**Answers with Solutions**

- (b)

x(jo) = x(t) e^{-jot} dt = e^{-2t}e^{-jot} dt

= e^{-3(2 + jo})/2 + j

- (a)

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

e^{-4|t|} = {e^{-4t}, t > 0 e^{4t} , t > 0

= e^{4t } e^{-jot} + e^{-4t} e^{-jot} dt

= 8/16 + 0^{2}

- (c)

x(t + t_{0}) – e^{jot0} x(j)

x (j) = (t + 1) + (t – 1)] e^{-jot} dt

= e^{jo} + e^{-jo} = 2 cos

- (c)

te^{-at} u(t) – 1/(a + j)^{2}

x(j) = te^{-t} e^{-jot}

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

- (d)

x(j) = a^{m} (t – m) e^{-jot} dt

= (ae^{-jo})^{m} = 1/1 – ae^{-jo}

- (c)

e^{-t} u(t) – 1/1 + j

x_{1}(t – 2) – e^{-2jo} x_{1} (j) [using shifting property]

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

- (b)

e^{-t} u(t) – 1/1 + j

e^{-3|t|} – 6/9 + 0^{2}

e^{j2t} x_{1}(t) – x_{1} {j(0 – 2)}

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

- (a)

e^{-3 |t|} = {e^{-3t} , t > 0 e^{3t} , t > 0

e^{-3|t|} – 6/9 + 0^{2}

x_{1} (t – 1) – e^{-jo} x_{1}(j)

tx_{2}(t) – jd/d x_{2} (j)

x (j) = j d/d [e^{-jo} 6/9 + 0^{2}] = 6e^{-jo}/9 + 0^{2} – 12 je^{-2jo}/(9 + 0^{2})

- (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

- (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

- (b)

e^{-a|t|} – 2a/a^{2} + 0^{2}

e^{-a|t|} = {e^{-at}, t > 0 e^{at}, t < 0

using duality property,

2a/a^{2} + t^{2} – 2e^{-a|0|} = 2e^{-a|0|}

1/a^{2} + r^{2} – a e^{-a|0|}

- (c)

x(j) = e^{t} 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 + 0^{2}

- (b)

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

- (c)

x(t) = 1/2 [2 (0) + (0 – 4) + (0 + 4)]e^{jot} d

= 1 + 1/2 e^{j4t} + 1/2 e^{-j4t} = 1 + cos 4t

- (a)

e^{-2|0|} = {e^{-20}, 0 > 0 e^{20}, 0 < 0

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

= 1/2 e^{2 }e^{jot} + 1/2 e^{-2} e^{-jot }d = 2/(4 + t^{2})

- (c)

x(e^{j}) = x[n] e^{-jn} = e^{-jn}

= e^{j2 }+ e^{j }+ 1 + e^{-j} + e^{-j2}

= e^{-j2 }(1 + e^{j }+ e^{j2} + e^{j3} + e^{j4})

= e^{-j2} (1 – e^{j5})/1 – e^{j}

= e^{j5/2} – e^{j5/2}/e^{-j/2} – e^{j/2} = sin 2.5 /sin 0.5

- (a)

x(e^{j}) = x[n]e^{-jn}

= (3/4) ne^{jn} = (3/4 e^{-j})^{n} = (3/4 e^{-j})/1 – 3/4 e^{-j}

- (c)

x[n] = u[n – 2] – u[n – 6]

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

x(e^{j}) = x[n] e^{jn}

= e^{-2j }+ e^{-3j} + e^{-4j} + e^{-5j}

- (b)

x(e^{j}) = a^{-n} e^{jn} + a^{n}e^{-jn}

= (ae^{j})^{-n} + (ae^{-j})^{n}

= ae^{j}/1 – ae^{j} + 1/1 – ae^{-j} = (1 – a^{2})/(1 + a^{2 }– 2a cos )

- (c)

x(e^{j}) = (1/2)^{-n} u[-n – 1] e^{jn}

= (1/2)^{-n} e^{jn} = (e^{-j}/2)^{-n}

= (e^{j}/2)^{n}

= 1/2 e^{j}/1 – 1/2 e^{j} = e^{j}/2 – e^{j}

- (a)

x(e^{j}) = 2[4 – 2n] e^{-jn} = 2e^{-j2}