i urgently need solution for this sinewave generator problem

Thread Starter

chintu84

Joined Oct 18, 2006
6
The circuit below has been designed as a simple sine wave generator, to produce an output voltage with an amplitude of 5.9 V at a frequency of 1 kHz. The low-pass filter has been designed to have a low-frequency gain of -1 and a cutoff frequency of 1.5 kHz. What is the magnitude of the undesired frequency component in the output waveform at 5 kHz ?
Express your answer in V, with an accuracy of 1 mV.

i attwched the circuit diagram.
i have tried this problem but i was not able get it.
i have got the wrong answer.
i have got it to be 0.2873.
but the actual answer is 0.081
i have tried it many times but i was not able to get it.
can anyone help me urgently.
 

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Thread Starter

chintu84

Joined Oct 18, 2006
6
The circuit below has been designed as a simple sine wave generator, to produce an output voltage with an amplitude of 5.9 V at a frequency of 1 kHz. The low-pass filter has been designed to have a low-frequency gain of -1 and a cutoff frequency of 1.5 kHz. What is the magnitude of the undesired frequency component in the output waveform at 5 kHz ?
Express your answer in V, with an accuracy of 1 mV.

i attwched the circuit diagram.
i have tried this problem but i was not able get it.
i have got the wrong answer.
i have got it to be 0.2873.
but the actual answer is 0.081
i have tried it many times but i was not able to get it.
can anyone help me urgently.
nothing was given except for this data.
can anyone help its urgent
 

Papabravo

Joined Feb 24, 2006
14,655
You cannot succeed in making this calculation until you can determine the order of the RC filter and what the slope of the rolloff is. That's the whole point of the problem.
 

Papabravo

Joined Feb 24, 2006
14,655
On a plot of gain versus frequency, the DC gain of -1 stands for 0 dB. The corner frequency of 1.5 kHz is where the actual response is 3 db down. It is from this point that you draw a straight line with a slope of 20 dB/decade. So the points are
(1.5kHz, 0 dB) and (15 kHz, -20 dB)
Now with a little bit of elementary algebra you should be able to figure out the attenuation at 5 kHz. Right?

Once you have the attenuation factor in dB you can compute the actual level.
 
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