# Digital Time Signal Processing

**Topics:**Finite impulse response, Digital signal processing, Window function

**Pages:**59 (7466 words)

**Published:**February 1, 2011

D T S P

DEC- 2004

By Kiran Talele ( talelesir@yahoo.com )

Q 1. (a) FIR filter described by the difference equation : y(n) = x (n) + x (n – 4) (i) Compute and sketch magnitude and phase response. [4] ⎛π ⎞ ⎛π ⎞ (ii) Find its response to the input x(n) = cos⎜ n ⎟ + cos⎜ n ⎟, − ∞ < n < ∞. ⎝2 ⎠ ⎝4 ⎠

[4] Solution : (i) To find Magnitude and Phase Response

Given (i) By ZT, y (n) = x (n) + x (n – 4) Y (z) = x (z) + z -4 x (z = x (z) (1 + z -4) H (z) = 1 + z -4 z = e jw H (e jw) = 1 + e –j4w

Put

= e − j2 w e j2 w + e − j2 w H (e jw ) = e − j2 w [2 cos (2w )] (i) (ii) (iii) w 0 0.1 π 0.2 π 0.3 π 0.4 π 0.5 π 0.6 π 0.7 π 0.8 π 0.9 π π

[

]

Magnitude Response M (w) = | Hr(w) | = | 2 cos (2w) | Phase Response : φ ( w ) = e − j2 w Phase : φ = − 2 w Hr(w) 2 1.62 0.62 – 0.62 – 1.62 –2 –1.62 – 0.62 0.62 1.62 2 Phase : φ 0 – 0.2 π – 0.4 π – 0.6 π + π = 0.4 π – 0.8 π + π = 0.2 π – π + π =0 – 1.2 π + π = – 0.2 π – 1.4 π + π = – 0.4 π – 1.6 π + 2 π = 0.4 π –1.8 π + 2 π = 0.2 π – 2 π + 2 π =0

1 2 3 4 5 6 7 8 9 10 11

DSP Help Line :

9987030881

www.guideforengineers.com

B E EXTC

DTS P

DEC- 2004

2

Magnitude Response | H(w)| 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.3 0.2 0 0.1π 0.2π 0.4π 0.6π 0.8π

w π

Phase Response

∅

0.4π 0.3π 0.2π 0.1π 0 –0.1π –0.2π –0.3π –0.4π

w

0

0.2π 0.4π 0.6π 0.8π π

Solution : (a) (ii) To find response : The frequency components present in the input signal x(n) are, w1 = π 2 π 4

and

w2 = π 2

w2 =

At

⎛ π⎞ M ( w) = 2 cos ⎜ 2 ⎟ = 2 cos (π ) = 2 ⎝ 2⎠ ⎛π⎞ φ = −2⎜ ⎟ + π = 0 ⎝2⎠

DSP Help Line :

9987030881

www.guideforengineers.com

B E EXTC

At

w2 = π 4

DTS P

DEC- 2004

3

π ⎛ π⎞ M ( w) = 2 cos ⎜ 2 ⎟ = 2 cos ( ) = 0 2 ⎝ 4⎠ π ⎛π⎞ φ = −2⎜ ⎟ = − 2 ⎝4⎠

The steady state response due to x(n) is then given by, y (n) = 2 cos ⎜ n ⎟ ANS ----------------------------------------------------------------------------------------------------------Q1. (b) Show that the zeros of a linear phase FIR filter occur at reciprocal locations. Also show that –– (i) FIR with symmetric impulse response and even length will compulsory have zero at z = –1. ⎛ π⎞ ⎝ 2⎠

(iii) (ii) FIR with anti-symmetric impulse response and odd length will compulsory have zero at z = + 1 and z = –1 [8] Solution : To find zeros of Linear phase FIR Filter. By ZT,

∞

H(z) =

∞

n =0

∑ h (n ) z −n

For linear phase FIR filter with symmetric h (n) ; h (n) = h ( N – 1– n)

H(z) =

n =0

∑ h ( N − 1 − n ) z −n

Put N – 1 – n = m m = N – 1- n

H( z) =

m =0

∑ h(m) z −( N−1−m)

m =0 − ( N −1)

∞

H(z) = z −( N −1)

∑ h (m) (z −1 ) −m

H(z −1 )

∞

H(z) = z

Let z1 be zero of the filter, Then H (z) z =z1 = 0 ie But

− H(z1 ) = z1−( N −1) H(z1 1 ) = 0 z1 ≠ 0 ( In FIR filter, POLES are always only at origin ) − H ( z1 1 ) = 0

Therefore, ie DSP Help Line :

H(z)

− z = z1 1

=0

9987030881

www.guideforengineers.com

B E EXTC

DTS P

=0

DEC- 2004

4

H(z) That means

z=

1 z1

1 is also zero of the filter. z1

When h[n] is symmetric or Antisymm. Filter is linear phase FIR filter and zeros of the filter are always in reciprocal order. Solution (b) (i) To find compulsory ZERO For FIR filter with symmetric h [n], ⎛1⎞ H (z) = z −( N −1) H⎜ ⎟ ⎝z⎠

When N is Even

⎛1⎞ H (z) = z −odd H⎜ ⎟ ⎝z⎠

At

Z = – 1,

H(−1) = (−1) −odd H(−1)

H ( −1) = − H ( −1) = 0

H (−1) = (−1) H (−1) H(z) z =−1 = 0 ie There exists definite zero at z = – 1.

Solution (b) (ii) To find compulsory ZERO

FIR filter with Anti-symmetric h [n], ⎛1⎞ H (z) = − z −( N −1) H⎜ ⎟ ⎝z⎠ When N is odd, ⎛1⎞ H (z) = − z − Even H⎜ ⎟ ⎝z⎠ a) At z = + 1, H (1) = – (1)Even H (1) H (1) = – H (1) = 0 H(z) z =1 = 0 ie There exists definite zero at z = 1. b) At z = – 1, ie H (1) = – ( –1)Even H (1) H (1) = – H ( –1) = 0 H(z) z=−1 = 0

There exists definite zero at z = – 1

DSP Help Line :

9987030881

www.guideforengineers.com

B E EXTC

DTS P...

Please join StudyMode to read the full document