# First order filter help

Discussion in 'Electronics Homework Help' started by cps13, Oct 24, 2021.

1. ### cps13

54
1
Feb 25, 2013
Hi,

I have the following question:

FIGURE 2 shows the response of a first-order filter to a step input. Estimate the half-power cut-off frequency of the filter and sketch its frequency and phase response.

My attempt at the answer is:

The half-power cut-off frequency is when the voltage is at 0.707 of maximum where the output is at -3dB. In this example, the output is a maximum of 1V and the output takes ~2.25us to reach 0.707 of this point. Therefore, the estimated cut-off frequency is:

2.25/10^-6 = 2.25MHz

Can somebody tell me if this is correct? Thanks.

Figure 2 is:  Moderators note : uploaded picture with better white balance for better readability.

Last edited by a moderator: Oct 31, 2021
2. ### Harald KappModeratorModerator

12,456
2,985
Nov 17, 2011
To the left you take the 2.25 (without adding the µs, which is also an omission), so the left side of the equation is in units of µs.
To the right you have a result in units of MHz which is 1/µs.
But in an equation both sides need to be equal, values as well as units. The mismatch shows that your equation can't be right.
See e.g. here for an explanation of the relation between bandwidth and rise time.

3. ### cps13

54
1
Feb 25, 2013
Hi - thanks for the reply but I don't follow. The reason the 2.25 is divided by 10^-6 is so that it is in uS and balanced to the MHz side.

4. ### Harald KappModeratorModerator

12,456
2,985
Nov 17, 2011
2.25/10^-6 = 2.25 × 106
no units at all, no µs, no MHz.
I assumed, maybe wrongly, that the 2.25 in your equation are the 2.25 µs from the point 2.25µs when the output reaches 0.707. In that case the equation would be 2.25 µs / 10-6 = 2.25 s, still not a frequency but a time.
MHz = 1/µs, not µs. That is where your equation fails. Read the text in the link I gave, it explains how to proceed.

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5. ### cps13

54
1
Feb 25, 2013
Looking at the section of your link called "What is an example of 3 dB bandwidth estimated from rise time?"

I believe that if I calculate my rise time, which is the time to go from a magnitude of 0.1 to 0.9, then I can relate this to my corner frequency (f3db). My rise time I calculate as:

8us - 1us = 7us

This is because the step change starts at 1us on the graph.

Therefore, using the formula from the material you shared:

Tr = 0.35/3fdb
7us = 0.35/3fdb
therefore

3fdb = 0.35 / 2.75us = 127,272Hz or 127kHz

Is this the correct way to calculate?

Thanks

6. ### Harald KappModeratorModerator

12,456
2,985
Nov 17, 2011
Your idea is basically correct, but how do you come from
7us = 0.35/3fdb
to
3fdb = 0.35 / 2.75us
Where do you take the 2.75 µs from when in the first equation you have 7µs?  