[SOLVED] Does Cut-off frequency of First order Low-Pass filter applies to Square wave?

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Hi,

We know the Cut-off frequency of First order Low Pass filter is, f = 1 / (2πRC).

And it means that the magnitude of Input freq. signal is attenuated to -3dB.

My understanding is that this applies only to a sinusoidal signal. And for square wave or other waves it has no meaning.

Can anybody tell whether my understanding is correct or not?
 

Filters are passive devices that like in your LPF case reflect any signal with a frequency above the cutoff. You should know that e.g. a square-wave signal is composed of harmonics. So the filter will cut off the harmonics except the first which is under the cutoff.
In practice it means that a suitably designed LPF will only pass the first harmonic, and the square-wave input signal will come out sinusoidal.
 

Of course, it depends on the repetition rate (frequency) of the square wave.

Let`s assume that the basic frequency of the square wave (first harmoinic) is identical to the cut-off frequency of the filter - it will be attenuated by 3dB.
All the other harmonics (at higher frequencies) will be attenuated more than 3dB - as detrmined by the filter`s ac response.
As a result (combination of all harmonics passing the filter) you will NOT have a sinusoidal wave because of the existence of the other harmonics.
Only in case of an ideal lowpass (not possible) the squarewave would be transfered to a pure sinus wave. In reality, a higher-order lowpass is sufficient in most cases.
 

The output of a first order filter can be most simply predicted in time domain, it's just a succession of exponential waves. The first derivative will be always discontinuous, so it can be clearly distinguished from a sine.
 

A square wave can be seen as sudden transitions to different DC levels. It is a different kind of instigator from a sinewave. It does not exhibit a phase shift.

You can only obtain a triangle-like output, by feeding a square wave to a first order low-pass filter.
 

A square wave consists of many component sinewaves, mainly of say V at the fundamental frequency F, then V/3 at 3F, V/5 at 5F, V/7 at 7F. This can continue ad infinitum, depending on the squarewave's rise time. The components are also in phase at some point in time.
So when you pass the above sinewaves through your filter, two things happen, the first is that the higher order harmonics get further attenuated, causing the rise time to increase, and secondly, the higher order sinewave progressively lag in time so the trailing edges of the square will show ringing.
Frank
 

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