Transfer function question

Discussion in 'General Electronics Chat' started by lkgan, Dec 21, 2009.

Dec 18, 2009
117
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Hi everyone,

Can anyone explain what this statement means?

A circuit having a transfer function Vout(s) / Vin(s) = 1 / (1 + s/ωo) is considered a low-pass filter because if Vin varies rapidly, Vout cannot fully track the input variations.

Thanks!

2. steinar96 Active Member

Apr 18, 2009
239
4
to keep it simple. The transfer function describes what the circuit does to the frequency components of a signal. The symbol used for angular velocity is ω and the relationship between the angular velocity and frequency of the circuit is f = 2*pi*ω. So as the frequency increases the term ω must increase right?

If you look at the given transfer function you ssee that ω appears in the denominator. So as ω increases the value of the transfer function decreases. This means that rapidly changing input signal (with high ω) result in a low output signal.

Dec 18, 2009
117
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Hi Steinar96,

Indeed the ω is in the denominator of (1 + s/ωo) term. If ω increase, the term will decrease. However, this term is the denominator of the whole transfer function. So eventually the transfer function will increase right?
And what I don't understand is that why it says when Vin varies rapidly, Vout cannot fully track the input variations. Is it mean when Vin change very fast, Vout cannot fully its changes?

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
It is the term in 's' that accounts for the input signal frequency parameter - not the (fixed) ω0 term. So the denominator does in fact increase with increasing frequency.

Consider what happens when a sinusoidal term at a certain frequency is applied as input to the transfer function. While the output will be a sinusoid of the same frequency, both the resulting amplitude and phase shift will differ from that of the input - depending upon the ratio of the input frequency to the fixed cut-off frequency ω0.

At ω=ω0 the amplitude will be reduced by a factor of 1/√2 and this is frequently referred to as the -3dB point or cut-off frequency. The phase of the output will be lagging that of the input by 45°. As the input frequency increases further beyond ω0, the output amplitude further decreases. The phase lag increases with respect to that of the input. This phase delay can also be viewed as a time shift or delay of the output in relation to the input.

Perhaps it is this phase lag (or time delay) that is the crux of the statement about the tracking of output in relation to input. Or perhaps the statement is simply making the point that output is not the same as input. Unfortunately while the intention in making such a statement might be to summarize the essential idea of the transfer function behaviour, it doesn't tell the whole story. Hence all these words of mine (useful or otherwise!) seeking to provide further clarification.