GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 42 of 65

Marks: +1, -0

It is desired to find a three-tap causal filter which gives zero signal as an output to an input of the form

x[n]=c_{1} \exp (-\;{j π n}/{2})+c_{2} \exp (\;{j π n}/{2})

where

{c}_{1}

and

{c}_{2}

are arbitrary real numbers. The desired three-tap filter is given by

{h}[0]=1, {h}[1]= {a}, {h}[2]= {b}

and

h[n]=0

for

n<0

n>2

What are the values of the filter taps a and

{b}

if the output is

{y}[ {n}]=0

for all

{n}

, when

{x}[ {n}]

is as given above?

Solution:

Given that

h(0)=1, h(1)=a, h(2)=b

and

h(n)=0

otherwise

∴ H(e^{j w})=1+a e^{-j w}+b e^{-j 2 w}

Now

{y}( {n})=0

for all

{n}

Now

x(n)=C_{1} e^{(\;{-j π n}/{2})}+C_{2} e^{(\;{j π n}/{2})}

If we consider

{C}_{1} {e}^{(\;{- {j} π {n}}/{2})}

as input then
Output

=C_{1}[1+a e^{+j \;{π}/{2}}+b e^{-j 2(-\;{π}/{2})}]

Output

= {C}_{1}[1+ {ae}^{ {j} \;{π}/{2}}+ {b} {e}^{ {j} π}]

........(i)
If we consider

{C}_{2} {e}^{(\;{ {j} π {n}}/{2})}

as input then
Output

=C_{2}[1+a e^{-\;{π}/{2}}+b e^{-j 2(\;{π}/{2})}]

=C_{2}[1+a e^{-j \;{π}/{2}}+b e^{-j π}]

......(ii)
Both output (i) and (ii) will be zero if

a=0, b=1

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