New answers tagged

0

Is it possible to design a digital filter to remove the ripple in the passband, Yes, but. In addition to the difficulty that Hilmar mentions, just pertaining to finding a filter, if you try this you'll find that a real analog filter also varies from part to part, and over temperature, aging, etc. So if your need is critical, and you're designing for ...


2

Is it possible to design a digital filter to remove the ripple in the passband Yes, but it's difficult and it also depends on the specific filter and the exact requirements of your application. Generally this can be done by designing either and IIR of FIR filter with a least square error optimization method using the inverse filter as a target. The tricky ...


2

Well, 514 is close enough to a power of two, when a floating point number crosses a power of two it looses one digit accuracy. The floating point number will be stored with 24-bit mantissa. Your sum will be around $2^{15}$, when the sum is less than $2^{15}$, your input will be accumulated with precision $2^{14-24}=2^{-10}$, if the sum accumulator is greater ...


2

Of course it's valid for real coefficients because real number is a subset of complex number and the proof doesn't make any assumption whether the coefficient is complex or real. For a complex $d$ $$ \begin{aligned} |A(z)|^2 &= A(z)A^*(z) = \frac{1-d^*z}{z-d} \frac{1-dz^*}{z^*-d^*}\\ &=\frac{1-(d^*z+dz^*)+|d|^2|z|^2}{|z|^2 - (d^*z+dz^*) + |d|^2} \end{...


0


1

Tim has it correct, it's a rounding error. You have a very low cutoff frequency, which means that your pole will extremely close to the unit circle (or $ z= 1$) in this case and you need to practice good numerical "hygiene" so to speak. Working with rounded numbers is not going to work here, you need "exact" within the limit of your ...


1

this is where things start to to awry Actually, no. Here's where things go sideways: h_lpf = 1.331e-05 z + 1.331e-05 ----------------------- z - 1 Sample time: 1.8824e-11 seconds Discrete-time transfer function. Or, rather, where you see what Matlab has printed out, and you take it for truth. It's pretty obvious that the z - 1 in your ...


1

The notion of a single decay time for an IIR filter can be a bit tricky. The decay times are solely a functions of the pole locations: the closer a pole to the unit circle, the longer the decay time is. In the early part of the impulse response, the decay of all poles is visible. However, the poles with lower decay time "die off" first and at the ...


1

Yes, you can design FIR filter of any finite order. You can re-write the examples by multiplying Z in the numerator and denominator to get: $$\frac{1}{2}\frac{(z+1)}{z}$$ Where it is clearer that you have only one zero and one pole. Therefore, this filter is of first order. In stable and causal FIR filters the order is given by the number of poles, and these ...


1

What you describe is a linear phase. Assuming that your sample rate is 48 kHz, you can implement this simply with a delay of -1.2 samples. The tricky parts here are that the delay is negative, i.e. the filter is non causal and that the delay is fractional (and not an integer number of samples). This can all be done, but needs to be carefully tailored to the ...


0

And they told that the measurement was taken according to the ISO 4869-1:1994 norm. But I'm unable to find the norm. The measurements references IEC 60318 which is available at their webstore. How these values in dB should be understand? The first set of numbers is the sensitivity in Pa/V (Pascal per Volt). That's the ratio of the sound pressure measured ...


Top 50 recent answers are included