> This is the part I'm having a tough time understanding. For a practical design, you need to prewarp. Is serious warping of significant frequencies as the upper limit isĪpproached. Moreover, since the BLT maps analog 0-oo to digital 0-fs/2, there (Mathematically, anyway.)Īnalod signals, and hence analog filters go to to fs/2. Prototype filters go from zero to infinity. No digital signal can represent frequencies outside that range. > filters that are restrained to that frequency range. This means that I'm restricted to designing Z = e^st is the definition of z, It applies to all digital difference Also I think I should clarify my question a 'f domain' means the frequency domain which resultsįrom applying the FT in continuous-time domain. Results from a DFT in discrete time domain, and Here 'w domain' means the frequency domain which This cancels the effect of theīLT in step 5) so that the end result is a So you can think of step 3) above as an 'inverse BLT' The recipe goes more or less as follows :Ģ) Convert the spec to an equivalent LP prototypeĥ) Use the BLT to transform back to w domainĦ) Convert from the LP prototype to the target filter The big picture: You want to design a discrete IIR filter Yet the functions asks u to provide something > That sounds to me like you should be able to design > transform theoretically transforms the entire frequency axis -inf to inf Hz That's the usual way of doing things, yes. > the BILINEAR transform to get it back into discrete domain. > first design the analog LP, then convert to the LP, HP, BP you want and use > help menu for ellip() and cheb2(), the description says that the functions This means that I'm restricted to designing > cheb2(), I'm asked to provide a normalized frequency from 0 to 1, > When I call on a digital filter design function such as ellip() or Also I think I should clarify my question a > Thanks for the reply! However I believe your definition of z=exp(sT) is >No digital filter can operate about fs/2. >So if you sample at 10kHz how can you possibly have a digital filter >unit circle forever and repeating the freq response you already have. >T=1/fs you will quickly see that "any frequency above fs/2" gives a >rational) Although it applies to any frequency, by substitution of >is a mapping of s to z via an approximation of z=exp(sT).(to make it >The max freq you can get for any sampled system is fs/2 or pi radians. > transform, which considers all frequencies from 0 to inf Hz. However, all digital filter functions involve the use of > In the description it says we are only considering frequencies from 0 > Ok, so I'm getting really confused by the way matlab is designing >On Oct 2, 6:57 am, "Fishilicious" wrote: Yet the functions asks u to provide something That sounds to me like you should be able to designĪnything from 0 to inf Hz. Transform theoretically transforms the entire frequency axis -inf to inf Hz This is the part I'm having a tough time understanding. The BILINEAR transform to get it back into discrete domain. Help menu for ellip() and cheb2(), the description says that the functionsįirst design the analog LP, then convert to the LP, HP, BP you want and use This means that I'm restricted to designingįilters that are restrained to that frequency range. When I call on a digital filter design function such as ellip() orĬheb2(), I'm asked to provide a normalized frequency from 0 to 1,Ĭorresponding to 0 to fs/2. Thanks for the reply! However I believe your definition of z=exp(sT) is No digital filter can operate about fs/2. So if you sample at 10kHz how can you possibly have a digital filter Unit circle forever and repeating the freq response you already have. T=1/fs you will quickly see that "any frequency above fs/2" gives a Rational) Although it applies to any frequency, by substitution of Is a mapping of s to z via an approximation of z=exp(sT).(to make it The max freq you can get for any sampled system is fs/2 or pi radians. > please clarify what is going on? Thanks! However, all digital filter functions involve the use of bilinear > In the description it says we are only considering frequencies from 0 to > Ok, so I'm getting really confused by the way matlab is designing filters.