# Is the magnitude function and phase function of its

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] is the magnitude function and phase function of its DTFT X ( ϖ ˆ j e ): Magnitude spectrum: 2 ˆ 2 ˆ ˆ )} ( Im{ )} ( Re{ | ) ( | ϖ ϖ ϖ j j j e X e X e X + = Magnitude spectrum in dB is 20log 10 | X ( ϖ ˆ j e )| Phase spectrum: } )} ( Re{ )} ( Im{ { tan ) ( ˆ ˆ 1 ˆ ϖ ϖ ϖ j j j e X e X e X - = Example: Exponential sequence with α =0.9 The DTFT of the exponential sequence is ) ˆ sin ˆ cos 1 ( 1 1 1 ) ( ˆ ˆ ϖ α ϖ α α ϖ ϖ j e e X j j + - = - = - The magnitude spectrum is ϖ α ϖ α ϖ ˆ sin ) ˆ cos 1 ( 1 | ) ( | 2 2 2 2 ˆ + - = j e X The phase spectrum is } ˆ cos 1 ˆ sin { tan ) ( 1 ˆ ϖ α ϖ α ϖ - - = - j e X

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-3 -2 -1 0 1 2 3 -10 -5 0 5 10 15 20 frequency magnitude spectrum in dB -3 -2 -1 0 1 2 3 0 2 4 6 8 10 12 frequency magnitude spectrum -3 -2 -1 0 1 2 3 -1.5 -1 -0.5 0 0.5 1 1.5 frequency phase spectrum Example: Consider an FIR filter with the impulse response coefficients { k b }={1, 2, 1}. The frequency response is given by ϖ ϖ ϖ ˆ 2 ˆ ˆ 2 1 ) ( j j j e e e H - - + + = Find the magnitude and phase of the frequency response:
As the coefficients are symmetric, we can obtain a simple amplitude function and a linear phase. We express the frequency response as follows: ) 2 ( 2 1 ) ( ˆ ˆ ˆ ˆ 2 ˆ ˆ ϖ ϖ ϖ ϖ ϖ ϖ j j j j j j e e e e e e H - - - - + + = + + = ) ˆ cos 2 2 ( ˆ ϖ ϖ + = - j e Since 0 ) ˆ cos 2 2 ( + ϖ for π ϖ π < - ˆ , the magnitude of the frequency response is ) ˆ cos( 2 2 ) ( ˆ ϖ ϖ + = j e H and the phase is ϖ ϖ ˆ ) ( ˆ - = j e H . Example: Consider an FIR system with the impulse response coefficients { k b }={1, - 1, 1}. The frequency response is given by

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ϖ ϖ ϖ ˆ 2 ˆ ˆ 1 ) ( j j j e e e H - - + - = Find the magnitude and phase of the frequency response: As the coefficients are symmetric, we can obtain a simple amplitude function and a linear phase. We express the frequency response as follows: ) 1 ( 1 ) ( ˆ ˆ ˆ ˆ 2 ˆ ˆ ϖ ϖ ϖ ϖ ϖ ϖ j j j j j j e e e e e e H - - - - + - = + - = ) 1 ˆ cos 2 ( ˆ - = - ϖ ϖ j e otherwise for e e j j 3 / | ˆ | ) ˆ cos 2 1 ( ) 1 ˆ cos 2 ( ) ˆ ( ˆ π ϖ ϖ ϖ π ϖ ϖ < - - = + - - -3 -2 -1 0 1 2 3 0.5 1 1.5 2 2.5 3 Magnitude response -3 -2 -1 0 1 2 3 -2 -1 0 1 2 normalized radian frequency Phase response The frequency response equals zero when 3 / ˆ π ϖ = . Properties of DTFT (i) Periodicity
The DTFT is periodic with a period of 2 π . That is ) ( ) ( ˆ ) 2 ˆ ( ϖ π ϖ j k j e X e X = + Please see the proof in the textbook. (ii) Linearity Since DTFT is an integral operation on x [ n ], DTFT is a linear transform. For any scalar a and b , and data sequences x [ n ] and y [ n ] with their DTFTs ) ( ˆ ϖ j e X and ) ( ˆ ϖ j e Y , we have ) ( ) ( ] [ ] [ ˆ ˆ ϖ ϖ j j e bY e aX n by n ax + + (iii) Symmetry If x [ n ] is a real-valued discrete-time signal, then its DTFT ha s conjugate symmetry, i.e. ) ( ) ( ˆ * ˆ ϖ ϖ j j e X e X - = or ) ( ) ( ˆ ˆ * ϖ ϖ j j e X e X - = Proof * ) ˆ ( * ˆ ˆ ˆ ) ] [ ( ) ] [ ( ] [ ) ( n j n n j n n j n j e n x e n x e n x e X ϖ ϖ ϖ ϖ - - -∞ = -∞ = - -∞ = = = = ) ( ˆ * ϖ j e X - = This result shows that )} ( Im{ )} ( Re{ )} ( Im{ )} ( Re{ ˆ ˆ ˆ ˆ ϖ ϖ ϖ ϖ j j j j e X j e X e X j e X - - - = + We have )} ( Re{ )} ( Re{ ˆ ˆ ϖ ϖ j j e X e X - = )} ( Im{ )} ( Im{ ˆ ˆ ϖ ϖ j j e X e X - - =

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)} ( Re{ ˆ ϖ j e X is an even function of ϖ ˆ )} ( Im{ ˆ ϖ j e X is an odd function of ϖ ˆ | ) ( | | ) ( | ˆ ˆ ϖ ϖ j j e X e X - = magnitude function is an even function of ϖ ˆ ) ( ) ( ˆ ˆ ϖ ϖ j j e X e X - -∠ =
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