The Electrician
- Joined Oct 9, 2007
- 2,970
I have solved your equations and they are correct. The solution for Vout is a messy expression involving Vin. You factor out the Vin term and what's left is your transfer function, which you could refer to as tfr(s).Sorry again but update:
Realised i got the S parameter. Went to convert it to f and i got only 3 positive values...
View attachment 162412
corresponding to the 1st, 4th and 5th value.
they correspond to 26 MHz, 28MHz and 24.5Mhz... not sure if its correct but finally see some values lying close to the actual frequency wanted...
Next you should produce a Log Log plot of the frequency response. You could plot versus radian frequqncy, or versus Hertz (f). To plot versus f , you need to make the substitution s = j 2 Pi f. Also, be aware that when you have done this, the transfer function will be a function of a complex variable, returning a complex result. You can't plot a complex result on a two dimensional plot.
Matlab should have a number of built-in functions for dealing with complex numbers, such as Im(), Re(), Abs(), Arg(), etc. You must plot the magnitude of the transfer function (which is a real number), not the transfer function itself (which is a complex number); actually you should plot the Log of the magnitude of the transfer function versus Log of the frequency. Let the transfer function be denoted by tfr(); then Abs(tfr()) is what you will use to get the magnitude of the transfer function.
When you produce the plot, you will see that the maximum magnitude is .5 A problem is that the response in the passband has some ripple, so you will have to determine what value to use for the passband magnitude (you might have to ask the professor), which will be multiplied by .707 to give the magnitude at the cutoff frequqncies, which you can solve for.