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[SOLVED] Analog circuit frequency response analysis with zero frequency

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greenjuice

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Hello.

For instance, our transfer function follows as :

H(s) = (1+s/Wz1)/(1+s/Wp1)(1+s/Wp2)

When we get zero frequency in analog amplifier design, we assume that transfer function is equal to zero at zero frequency(because numerator will be zero)

from that, we can get wz1, zero frequency. it means s of s/Wz1 is zero in only real part. right?

But in the vantage point of signal, we substitute s=jw.

and at zero frequency we have +3dB point compared with previous amplitude. In summary, at zero frequency, it has some gain even if we get zero frequency under assumption that our H(s) = 0.

why do i have this gap? is it from analysis of system and signal?

actually I am very confused about that.

please help me.

Thank you
 

H(s) = (1+s/Wz1)/(1+s/Wp1)(1+s/Wp2)

When we get zero frequency in analog amplifier design, we assume that transfer function is equal to zero at zero frequency(because numerator will be zero)
Click to expand...
But H(0) of the above transfer function is actually unity for any finite pole and zero values.
 

No matter what the numerator, divide by zero never works
out (except as a trapped-error special case).

Use a minuscule real frequency, lower than the lowest you
care about, and call it good enough.
 

FvM said:
But H(0) of the above transfer function is actually unity for any finite pole and zero values.
Click to expand...

Oh i'm sorry for you to misunderstand my question. Zero frequency which i mean is zero. Not dc value or unity gain frequency. I'm sorry.
 

DC is zero frequency. Don't understand what you mean.
 

I cannot understand what you mean. Cna you explain in detail?

- - - Updated - - -

FvM said:
DC is zero frequency. Don't understand what you mean.
Click to expand...

It's my mistake.. what i really want to know is about zero and pole. Especially zero. For instance When we use zero pole cancellation technique in analog circuit we can get zero under assumption that vout =0( it means h(s)=0). But actually when we draw bode plot, at zero its gain is not 0 but some point where there are some gain. +3db point(in the case of pole -3db point as you know)

In conclusion my assumption and resulf of bode plot collide each other. But it's really true from analog circuit design written by razavi.

Please let me know. How can i organize my thought?
 

Your example transfer function has only real poles and zeros, means the denominator and numerator polynominal becomes zero for real values of s = σ + jω. But the bode diagram shows only stationary solutions with σ=0 respectively purely imaginary s.

See "Understanding Poles and Zeros" https://web.mit.edu/2.14/www/Handouts/PoleZero.pdf
 

    V

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Your statement "When we get zero frequency in analog amplifier design, we assume that transfer function is equal to zero at zero frequency(because numerator will be zero)" is probably not correct.

To my understanding, at zero frequency, the amplitude of the real part and the imaginary part are equal. However, the real part and the imaginary part have 90 degree phase difference. As a result, you will see a 3-dB gain increment in the Bode plot, and a 45 degree phase leading.

I hope I get your question correctly.
 

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