Let Xej- be the DTFT of x-n- where Xej- - 11-e-j- then the magnitude of DTFT of h-n- - -1n x-n- at - - -2 is

Given, x[n] DTFT⟷X(e^jω) h[n] = ( 1)^n x[n] = e^jωx[n] ∴ ( 1)^n = e^jω By taking DTFT, H(e^jω) = X(e^j(ω π)) =11 e^j(ω π) = 11 e^jω. e^ jπ H(e^jω) = 11+e^ ...

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Standard XII

Physics

Towards Biot-Savart Law

Let Xejω be t...

Question

$Let X (e^{j ω}) be the DTFT of x [n] where X (e^{j ω}) = \frac{1}{1 - (e^{- j ω})} then the magnitude of DTFT of h [n] = (- 1)^{n} x [n] at ω = \frac{π}{2} is ________$
0.707

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Solution

The correct option is A 0.707
Given, $x [n] D T F T ⟷ X (e^{j ω})$

$h [n] = (- 1)^{n} x [n]$

$= e^{j ω} x [n]$ $∴ (- 1)^{n} = e^{j ω}$

By taking DTFT, $H (e^{j ω}) = X (e^{j (ω - π)})$

$= \frac{1}{1 - e^{j (ω - π)}}$

$= \frac{1}{1 - e^{j ω} . e^{- j π}}$

$H (e^{j ω}) = \frac{1}{1 + e^{j ω}}$

$∴ e^{j π} = - 1$

at $ω = \frac{π}{2};$

$H (e^{j π / 2}) = \frac{1}{1 + e^{j π / 2}}$

$| H (e^{j π / 2}) | = | \frac{1}{1 + j 1} | = \frac{1}{\sqrt{2}} = 0.707$

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Let X(ejω) be the DTFT of x[n] where X(ejω)=11−(e−jω) then the magnitude of DTFT of h[n]=(−1)nx[n] at ω=π2 is ________ 0.707

The correct option is A 0.707Given, x[n]DTFT⟷X(ejω) h[n]=(−1)nx[n] =ejωx[n] ∴(−1)n=ejω By taking DTFT, H(ejω)=X(ej(ω−π)) =11−ej(ω−π) =11−ejω.e−jπ H(ejω)=11+ejω ∴ejπ=−1 at ω=π2; H(ejπ/2)=11+ejπ/2 |H(ejπ/2)|=|11+j1|=1√2=0.707

$Let X (e^{j ω}) be the DTFT of x [n] where X (e^{j ω}) = \frac{1}{1 - (e^{- j ω})} then the magnitude of DTFT of h [n] = (- 1)^{n} x [n] at ω = \frac{π}{2} is ________$
0.707