Chap6StudentSolutions

In the frequency domain 1 m v2 f 25 sinc sinc vi

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Unformatted text preview: on of the original signal. (This is a simplified model but it illustrates the essential features of a chopper-stabilized amplifier.) Let the following be the parameters of the chopper-stabilized amplifier: Chopping frequency Gain of the first amplifier Bandpass filter Gain of the second amplifier Lowpass filter 500 Hz 100 V/V Unity-gain, ideal, zero-phase. Passband 250 < f < 750 10 V/V Unity-gain, ideal, zero-phase. Bandwidth 100 Hz Let the excitation signal have a 100 Hz bandwidth. What is the effective DC gain of this chopper-stabilized amplifier? Let the excitation be v i ( t) . Then the signal after the first amplifier is v1 ( t) = 100 v i ( t)[rect (1000 t) ∗ 500 comb(500 t)] or, in the frequency domain, V1 ( f ) = 50 ∞ m ∑ sinc 2 V ( f − 500m) m =−∞ i The response signal from the bandpass filter is that part of the spectrum lying between 250 and 750 Hz which is 1 Vbpf ( f ) = 50 sinc [Vi ( f − 500) + Vi ( f + 500)] . 2 Solutions 6-40 M. J. Roberts - 8/16/04 The response of the second amplifier is the response of the BPF pulse-amplitude modulated by the synchronous switching. In the frequency domain, ∞ 1 m V2 ( f ) = 25 sinc ∑ sinc [Vi ( f − 500 m − 500) + Vi ( f − 500 m + 500)] 2 m =−∞ 2 If the excitation is dc, then Vi ( f ) = Aδ ( f ) and ∞ 1 m V2 ( f ) = 25 A sinc ∑ sinc [δ ( f − 500 m − 500) + δ ( f − 500 m + 500)] . 2 m =−∞ 2 The r...
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