Electrical Engineering - DSP analogue signal question?

Question

could someone please help me with the following question

On the NanoBoard the anti-aliasing and reconstruction filters are passive low pass filters both with nominal cut-off frequencies around

10 kHz. The Maxim MAX1104 is a full codec and so contains both analogue to digital (A2D) and digital to analogue (D2A) converters, these being represented by the sample and hold (S/H) plus quantiser (Q) and D/A blocks in Figure 1 respectively. Finally, the DSP block in Figure 1 is where the difference equation for implementing the desired digital filter is performed. figure 1 is a block diagram of the following

x(t)->anti-aliasing filter->S/H->Q->DSP->D/A->reconstruction filter->y(t)

Could someone help me with these questions questions:

1. Bearing in mind that the anti-aliasing filter on the NB1 is not ideal, explain what happens to the analogue signal, x(t), after it has passed through this filter and is presented to the input of the A2D,

when:

a. x(t) is a sinusoidal signal of frequency 10 kHz.

b. x(t) is a square wave (50% duty cycle) of fundamental frequency 4 kHz. Hint, think about what the Fourier series representation of this square wave signal is.

Any help would be appreciated, Thanks.

Answer 1

a)  x(t) will be attenuated by the filter by about 3dB (as thats the cutoff point of the filter and its a pure sinusoid)

b)  A square wave is given by odd harmonics in the fourier series, so our square wave will no longer be a square wave coming out of the anti-aliasing filter. Anything higher than ~10kHz will be attenuated by the filter so you will end up with a 4kHz sine-wave looking signal with a  8kHz component in there too.

Answer 2

Something amounting to the same question was asked sometime in the last day or so.  If that was not you, take a look.

If it was you, then the only new things seems to be that the filters are non-ideal.  If they are non-ideal they can likely be approximated by a rolloff of x dB per octave, or similar, which will likely get you a close enough answer when you apply it to the Fourier harmonic decomposition of the squarewave. eg fs + 1/3 3fs + 1/5 5fs + ...

Question was

Electrical Engineering - reconstruction filter, what is the frequency and output?

But it seemed to be asked twice before.

Maybe he didn't like an answer that required some thought.

Answer 3

For the square wave, you need to know the attenuation curve of the filter (no of poles).  A single-pole filter will have an attenuation of -6db/octave.  The first harmonic of the square wave is at 12 kHz (n=3) and has an amplitude of 4/(3π) times the square wave amplitude.  The next harmonic (n=5) is at 20 kHz and has an amplitude of 4/(5π).  The first octave above the filter cutoff is 20 kHz, so the attenuation of the second harmonic is 6 db.  The attenuation of the first harmonic is less than 3 db.  The resulting wave will be smoothed somewhat because of the reduction in the amplitude of the higher harmonics.