-Tracking loop filter is a well-known concept in control theory. the tracking loop filter in GNSS RX is designed based on its analogous characteristics to control systems especially tracking stage. To have deeply understanding the tracking loop filter, we have to read them on control system, that would be showed on the reference part of this post.
- The aim of tracking filter is to estimated the reference parameter as more accurate as possible. In case of noise-free signal, those parameter can be estimated directly. Unfortunately, the incoming signal always includes noise. This leads to the use of loop filter in order to reduce the effect of error.
- To analyze the characteristic of a tracking loop filter, we must consider its effect on desired signal and noise separately. In most case, noise is assumed as white Gaussian noise.
2. Tracking loop filter design
We consider 2 kinds of input:
- Step input
- Ramp input
And the noise is assumed as white Gaussian noise
That leads to 2 kinds of filters namely: first-order and second-order filters
$$\hat{\xi}[k+1]=\hat{\xi}[k]+\epsilon_\xi[k]$$
$$\Rightarrow{D(z)}=\frac{1}{z-1}$$
2.1. Noise-free signal
We consider the behavior of tracking filer with 2 kinds of input:
Step input $\xi[k]=Au[k]$ ($\xi(z)=\frac{Az}{z-1}$) and ramp input $\xi[k]=Aku[k]$ ($\xi(z)=\frac{Az}{(z-1)^2}$)
We expect our estimate convergence to the true value. Moreover, the final value theorem states as following expressions:
$$\lim_{k\rightarrow\infty}f[k]=\lim_{s\rightarrow0}sf(s)=\lim_{z\rightarrow1}(z-1)f(z)$$
It's clear that:
For first-order LP, $F(z)=\gamma$
$$\lim_{z\rightarrow1}(z-1)\xi_{step}(z)=0$$
$$\lim_{z\rightarrow1}(z-1)\xi_{ramp}(z)=\frac{A}{\beta\gamma}$$
For second-order LP, $F(z)=\frac{az+b}{z-1}$
$$\lim_{z\rightarrow1}(z-1)\xi_{step}(z)=0$$
$$\lim_{z\rightarrow1}(z-1)\xi_{ramp}(z)=0$$
2.2. Signal plus noise
We consider only variance (jitter) at the output of the filter:
$$\xi_\eta[k]=\eta[k]*h[k]=\sum_{n=0}^{+\infty}h[n]\eta[k-n]$$
$$E(|\xi_\eta[k]|^2)=E(\sum_{n=0}^{+\infty}h[n]\eta[k-n]\sum_{p=0}^{+\infty}h^*[p]\eta^*[k-p])$$
$$=E(\sum_{n=0}^{+\infty}\sum_{p=0}^{+\infty}h[n]\eta[k-n]h^*[p]\eta^*[k-p])$$
$$=\sum_{n=0}^{+\infty}|h[n]|^2E(|\eta[n-k]|^2)=\sum_{n=0}^{+\infty}|h[n]|\sigma^2_\eta$$
$$=\int_{-0.5}^{0.5}|H(j2{\pi}f)|^2{df}$$
$B_{eq}$ is defined as follows:
$H_{eq}(f_a)=H(0)$ where $|f_a|\leq{B_{eq}}$
The bandwidth controls amount of noise allowed in the filter.
$$\int_{-0.5}^{0.5}|H(f)|^2df=T_s\int_{-F_s/2}^{F_s/2}H(f_a)df_a=T_s2B_{eq}H(0)$$
$$\int_{-0.5}^{0.5}|H(f)|^2df=T_s\int_{-F_s/2}^{F_s/2}H(f_a)df_a=T_s2B_{eq}|H(0)|^2$$
$$B_eq=\frac{\int_{-F_s/2}^{F_s/2}H(f_a)df_a}{2|H(0)|^2}$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\int_CH(z)H^*(z^{-1})\frac1zdz$$
$$g(z)=H(z)H^*(z^{-1})\frac1z$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\sum_kR_g(z_p)$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\sum_k\lim_{z\rightarrow{p_k}}(z-p_k)g(p_k)$$
$p_k$ is a pole lie inside the unitary circle C
$$|H(0)|^2=(H(z)H^*(z^{-1}))|z=1$$
a, First-order filter
$$|H(0)|=\frac{1}{\beta^2}$$
$g(z)$ have 2 poles inside C which are $z=0$ and $z=1-\beta\gamma$,
$$\sum_k\lim_{z\rightarrow{p_k}}(z-p_k)g(p_k)=\frac{\gamma^2}{1-(1-\beta\gamma)^2}$$
$$B_{eq}\approx\frac{\beta\gamma}{4T_s}$$
Finally:
$$\sigma^2_\xi\approx\sigma^2_\eta\frac{\gamma}{2\beta}$$
b, Second-order filter
For the filter, we have
$$|H(0)|=\frac{1}{\beta^2}$$
$$H_\eta(z)=\frac{-az+b}{(z-1)^2+\beta(az-b)}$$
$$g(z)=\frac{H(z)H^*(z^{-1})}{z}$$
If $H(z)$ have two poles namely $\alpha$ and $\alpha^*$, $H^*(z^{-1})$ have two poles $\frac{1}{\alpha}$ and $\frac{1}{\alpha^*}$. It's obvious that only two of them lie inside the unit circle. Suppose that $\alpha$ and its conjugate $\alpha^*$
(It's easy to prove that: $\alpha=1-\frac{{\beta}a}{2}+j\sqrt{1-{\beta}b-(1-\frac{{\beta}a}{2})^2}$)
$$B_{eq}=\frac{\beta^2}{T_s}R\{\frac{(a\alpha-b)(a^*-b^*\alpha)}{2jI(\alpha)(1-|\alpha|^2)(1-\alpha^2)}\}$$
$$\sigma_\delta=2\sigma_{\eta}R\{\frac{(a\alpha-b)(a^*-b^*\alpha)}{2jI(\alpha)(1-|\alpha|^2)(1-\alpha^2)}\}$$
The parameters a, b is chosen according to the analogous filter as follows:
$$\rho\approx\frac{4B_{eq}T_s}{1+4\zeta^2(1+2B_{eq}T_s)}$$
$${\beta}b\approx4\zeta^2\rho$$
$${\beta}a\approx4\zeta^2\rho(1+2\zeta^2\rho)$$
In practice damping factor is chosen: $\zeta=sqrt(2)$
As the formula above, jitter depends only the characteristic of filter
2. Tracking loop filter design
We consider 2 kinds of input:
- Step input
- Ramp input
And the noise is assumed as white Gaussian noise
That leads to 2 kinds of filters namely: first-order and second-order filters
a, Signal combination
The input signal $y[k,\xi]$ is combined with the reference signal $y_{ref}[k,\hat{\xi}[k]]$. The local reference signal have the same basic structure to input signal. We try to make a 'copy version' of the input signal but we don't know exactly the parameter $\xi$. Consequently, we have to design a system to estimate that parameter.
b, Discrimination function
The objective of the function is to extract the error $\hat{\xi}[k]-\xi$ from the combined signal $z[k,\xi]$
c, Low-pass loop filter
Unfortunately the input signal includes not only the desired signal but also noise. The filter reduces noise in order to get the more accurate estimate of $\xi$
d, Parameter estimation
The estimated parameter $\hat{\xi}$ must be updated every loop with the difference between input and local signal.
$$\hat{\xi}[k+1]=\hat{\xi}[k]+\epsilon_{\xi}[k]$$
e, Delay
The block characterize the behavior of the estimate will be utilized in next loop
f, Reference signal generator
Generate the signal
The linearized model can be expressed as following:
$$\hat{\xi}[z]=(\beta\delta_{\xi}[z]+\eta[z])F(z)D(z)$$
$$\delta_{\xi}[z]=\xi[z]-\hat{\xi}[z]$$
$$\Rightarrow\hat{\xi}[z]=\frac{{\beta}F(z)D(z)}{1+{\beta}F(z)D(z)}\xi[z]+\frac{F(z)D(z)}{1+{\beta}F(z)D(z)}\eta[z]$$
$$\Rightarrow\delta_{\xi}[z]=\frac{1}{1+{\beta}F(z)D(z)}\xi[z]-\frac{F(z)D(z)}{1+{\beta}F(z)D(z)}\eta[z]$$
Transfer functions from $\xi[k]$ and $\eta[k]$ to $\hat{\xi[k]}$
$$G_\xi(z)=\frac{{\beta}F(z)D(z)}{1+{\beta}F(z)D(z)}$$
$$G_\eta(z)=\frac{F(z)D(z)}{1+{\beta}F(z)D(z)}$$
Transfer functions from $\xi[k]$ and $\eta[k]$ to $\delta{\xi[k]}$
$$H_\xi(z)=\frac{1}{1+{\beta}F(z)D(z)}$$
$$H_\eta(z)=-\frac{F(z)D(z)}{1+{\beta}F(z)D(z)}$$
It's easy to indicate the function D(z)$$\hat{\xi}[k+1]=\hat{\xi}[k]+\epsilon_\xi[k]$$
$$\Rightarrow{D(z)}=\frac{1}{z-1}$$
2.1. Noise-free signal
We consider the behavior of tracking filer with 2 kinds of input:
Step input $\xi[k]=Au[k]$ ($\xi(z)=\frac{Az}{z-1}$) and ramp input $\xi[k]=Aku[k]$ ($\xi(z)=\frac{Az}{(z-1)^2}$)
We expect our estimate convergence to the true value. Moreover, the final value theorem states as following expressions:
$$\lim_{k\rightarrow\infty}f[k]=\lim_{s\rightarrow0}sf(s)=\lim_{z\rightarrow1}(z-1)f(z)$$
It's clear that:
For first-order LP, $F(z)=\gamma$
$$\lim_{z\rightarrow1}(z-1)\xi_{step}(z)=0$$
$$\lim_{z\rightarrow1}(z-1)\xi_{ramp}(z)=\frac{A}{\beta\gamma}$$
For second-order LP, $F(z)=\frac{az+b}{z-1}$
$$\lim_{z\rightarrow1}(z-1)\xi_{step}(z)=0$$
$$\lim_{z\rightarrow1}(z-1)\xi_{ramp}(z)=0$$
2.2. Signal plus noise
We consider only variance (jitter) at the output of the filter:
$$\xi_\eta[k]=\eta[k]*h[k]=\sum_{n=0}^{+\infty}h[n]\eta[k-n]$$
$$E(|\xi_\eta[k]|^2)=E(\sum_{n=0}^{+\infty}h[n]\eta[k-n]\sum_{p=0}^{+\infty}h^*[p]\eta^*[k-p])$$
$$=E(\sum_{n=0}^{+\infty}\sum_{p=0}^{+\infty}h[n]\eta[k-n]h^*[p]\eta^*[k-p])$$
$$=\sum_{n=0}^{+\infty}|h[n]|^2E(|\eta[n-k]|^2)=\sum_{n=0}^{+\infty}|h[n]|\sigma^2_\eta$$
$$=\int_{-0.5}^{0.5}|H(j2{\pi}f)|^2{df}$$
$B_{eq}$ is defined as follows:
$H_{eq}(f_a)=H(0)$ where $|f_a|\leq{B_{eq}}$
The bandwidth controls amount of noise allowed in the filter.
$$\int_{-0.5}^{0.5}|H(f)|^2df=T_s\int_{-F_s/2}^{F_s/2}H(f_a)df_a=T_s2B_{eq}H(0)$$
$$\int_{-0.5}^{0.5}|H(f)|^2df=T_s\int_{-F_s/2}^{F_s/2}H(f_a)df_a=T_s2B_{eq}|H(0)|^2$$
$$B_eq=\frac{\int_{-F_s/2}^{F_s/2}H(f_a)df_a}{2|H(0)|^2}$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\int_CH(z)H^*(z^{-1})\frac1zdz$$
$$g(z)=H(z)H^*(z^{-1})\frac1z$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\sum_kR_g(z_p)$$
$$=\frac{1}{2|H(0)|^2(j2\pi{T_s})}\sum_k\lim_{z\rightarrow{p_k}}(z-p_k)g(p_k)$$
$p_k$ is a pole lie inside the unitary circle C
$$|H(0)|^2=(H(z)H^*(z^{-1}))|z=1$$
a, First-order filter
$$|H(0)|=\frac{1}{\beta^2}$$
$g(z)$ have 2 poles inside C which are $z=0$ and $z=1-\beta\gamma$,
$$\sum_k\lim_{z\rightarrow{p_k}}(z-p_k)g(p_k)=\frac{\gamma^2}{1-(1-\beta\gamma)^2}$$
$$B_{eq}\approx\frac{\beta\gamma}{4T_s}$$
Finally:
$$\sigma^2_\xi\approx\sigma^2_\eta\frac{\gamma}{2\beta}$$
b, Second-order filter
For the filter, we have
$$|H(0)|=\frac{1}{\beta^2}$$
$$H_\eta(z)=\frac{-az+b}{(z-1)^2+\beta(az-b)}$$
$$g(z)=\frac{H(z)H^*(z^{-1})}{z}$$
If $H(z)$ have two poles namely $\alpha$ and $\alpha^*$, $H^*(z^{-1})$ have two poles $\frac{1}{\alpha}$ and $\frac{1}{\alpha^*}$. It's obvious that only two of them lie inside the unit circle. Suppose that $\alpha$ and its conjugate $\alpha^*$
(It's easy to prove that: $\alpha=1-\frac{{\beta}a}{2}+j\sqrt{1-{\beta}b-(1-\frac{{\beta}a}{2})^2}$)
$$B_{eq}=\frac{\beta^2}{T_s}R\{\frac{(a\alpha-b)(a^*-b^*\alpha)}{2jI(\alpha)(1-|\alpha|^2)(1-\alpha^2)}\}$$
$$\sigma_\delta=2\sigma_{\eta}R\{\frac{(a\alpha-b)(a^*-b^*\alpha)}{2jI(\alpha)(1-|\alpha|^2)(1-\alpha^2)}\}$$
The parameters a, b is chosen according to the analogous filter as follows:
$$\rho\approx\frac{4B_{eq}T_s}{1+4\zeta^2(1+2B_{eq}T_s)}$$
$${\beta}b\approx4\zeta^2\rho$$
$${\beta}a\approx4\zeta^2\rho(1+2\zeta^2\rho)$$
In practice damping factor is chosen: $\zeta=sqrt(2)$
As the formula above, jitter depends only the characteristic of filter
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