Introduction
When we process an audio signal, for example, adding a sound effect, sometimes we have to think about what is the effect of that processing on the audio? Does it raise the low frequencies or the high frequencies? Or does the frequency change periodically?
Of course, you can use your ears, but it’s too hard to do that. Not everyone has golden ears. Even if your ear is good enough to pick up the most subtle distinctions, how do you describe them in terms that humans can understand? After all, not everyone has the ability to describe things.
In summary, we need a quantifiable way to describe the process of processing. Needless to say, we can only describe this complex process mathematically. In this article, we will look at the “processing” process from multiple perspectives.
Some knowledge foundation
The following are some basic signal processing knowledge, which will not be very detailed, but all point to point, the purpose is to string together the whole content, forming a self-closed loop. More in-depth introduction, we are interested in the DSP book to see it.
Signal said
First, how do you represent an audio signal (we’re only talking about discrete signals)? Very simply, we describe it as an array, a signal of length N, which can be expressed as:
We call x[n] a discrete time series
Unit pulse signal
The mathematics of it looks like this:
It looks like this:
Linear time invariant filter
The so-called filter, in fact, is the signal “operation” : a signal, through the filter, into another signal. We focus here on Linearity and time-invariance; LTI) filter. LTI filters are widely used and are the most powerful family of filters. We can express an LTI filter with the difference equation:
Where x is the input sequence and y is the output sequence.
Impulse response
The result obtained by feeding the unit impulse signal into an LTI filter is called the impulse response and is usually expressed as
convolution
signal
Z transform
Now let’s talk about the Z transform, which is very important.
Z transform can transform discrete time series into expressions in complex frequency domain. Er… Well, the preceding sentence is copied by Baidu Encyclopedia, for those of you who have never been exposed to Z transformation, this sentence is simply too good. Let’s try to understand it from an emotional point of view, using concrete examples.
The formula for Z transformation is very simple, as follows:
For example, the discrete time series is: x[0]=1, x[1]=2, x[2]=3, then:
By Z transformation, we convert a sequence of N values into an expression of Z (a value). The Z transformation can be viewed as a linear transformation, as in the example above, and can be rewritten as a multiplication of two vectors:
There’s an important property about the Z transform, convolution. Two signals convolved in the time domain is equal to multiplying in the Z transform domain
As mentioned earlier, the signal passing through the filter can be expressed as a convolution process (4), then for
From the point of view of this formula,
The formula above also shows that the impulse response
There are a few other properties we need to know about the Z transform:
Therefore, for an LTI filter (3), we perform Z transformation to obtain:
By combining the same items, we get:
if
The transfer equation is obtained
Describe a filter from multiple angles
The Difference equation
As shown in equation (3), this is the difference equation. Let’s give two examples to get a feel for it.
and
An impulse signal is input into (14) to obtain the impulse response
An impulse signal is input into (15) to obtain the impulse response
Block diagram
Sometimes it is more consistent with human cognitive processes to display filters in graphical form. Take Equation (3) for example, its block diagram is as follows:
Well, it looks complicated. Let’s look at a simple example, a delayed signal:
It looks like this:
Impulse response
As mentioned earlier, an impulse signal (equation (3)) is fed into a filter and the output is observed to obtain the impulse response.
For example, there is a filter:
You can plot the impulse response of the filter using the following MATLAB code
g1 = 0.5^3; B = [1 0 0 g1]; % Feedforward coeffs
g2 = 0.9^5; A = [1 0 0 0 0 g2]; % Feedback coefficients
h = filter(B,A,[1,zeros(1.50)]); % Impulse response
% h = impz(B,A,50); % alternative in octave-forge or MSPTB
% Matlab-compatible plot:
clf; figure(1); stem([0:50],h,'-k'); axis([0 50 0.8 1.1]);
ylabel('Amplitude'); xlabel('Time (samples)'); grid;
Copy the code
Of course, there are more precise calculations
Z transformation is performed on Formula (16) to obtain the transfer equation:
take
Then we compute H(z) in the code, which is also 0.9972, exactly the same.
g1 = 0.5^3; B = [1 0 0 g1]; % Feedforward coeffs
g2 = 0.9^5; A = [1 0 0 0 0 g2]; % Feedback coefficients
h = filter(B,A,[1,zeros(1.50)]); % Impulse response
z_value = 2; % z = 2
z = zeros(1.length(h)) + z_value;
Z = z.^-(0:length(h)- 1);
HZ = h*Z'; % calculate H(z), the result is 0.9972
Copy the codeTransfer function
The transfer equation is already up here, right
Poles, zeros gain and gain Poles
Let’s multiply the top and the bottom of equation 13
We call
We call
We call
We plot all poles and zeros on a complex plane. If all poles are in the unit circle, then the filter is stable; If all zeros are in the unit circle, then we call this filter a minimum phase filter.
Frequency Response
The spectral response refers to the change in the frequency domain of the signal after it passes through the filter. It is defined as the spectrum of the output signal divided by the spectrum of the input signal
The spectral information of a signal can be obtained by Fourier transform (DTFT) :
Does DTFT look a lot like the Z transform to you? Formula (5)
So the spectral response is essentially
Among them
Spectrum response is divided into Amplitude response and Phase response.
The amplitude response is essentially
And the phase response is essentially
Practical examples
A thousand words are not as good as an actual example. Suppose we have a filter like this:
Z transform:
Calculate the transfer equation:
Factor it
Get two poles:
You get two zeros, and they’re both at
when
w = 0:0.001:pi;
E = exp(1j*w);
Hz = (4*(E.^2) - 4*E + 1) ./ ((E.^2) + E + 0.5);
amplitude_Hz = abs(Hz);
phase_Hz = angle(Hz);
figure(1);
plot(20*log10(amplitude_Hz));
figure(2);
plot(phase_Hz*180/pi);
Copy the codeThe result might look something like this:
As can be seen from the figure, this filter suppresses low frequency information in the signal and gains high frequency information. in
conclusion
This article describes several ways to recognize a filter,
- Difference equation
- Block diagram
- Impulse response
- Transfer equation
- Poles, zeros, and gains
- Spectral response
And enumerate relevant examples, let everybody have more in-depth understanding and experience.
Reproduced please indicate the source juejin.cn/post/684490…