Documentation

# iqcoef2imbal

Convert compensator coefficient to amplitude and phase imbalance

## Description

example

[A,P] = iqcoef2imbal(C) converts compensator coefficient C to its equivalent amplitude and phase imbalance.

## Examples

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Use iqcoef2imbal to estimate the amplitude and phase imbalance for a given complex coefficient. The coefficients are an output from the step function of the IQImbalanceCompensator.

Create a raised cosine transmit filter to generate a 64-QAM signal.

M = 64;
txFilt = comm.RaisedCosineTransmitFilter;

Modulate and filter random 64-ary symbols.

data = randi([0 M-1],100000,1);
dataMod = qammod(data,M);
txSig = step(txFilt,dataMod);

Specify amplitude and phase imbalance.

ampImb = 2; % dB
phImb = 15; % degrees

Apply the specified I/Q imbalance.

gainI = 10.^(0.5*ampImb/20);
gainQ = 10.^(-0.5*ampImb/20);
imbI = real(txSig)*gainI*exp(-0.5i*phImb*pi/180);
imbQ = imag(txSig)*gainQ*exp(1i*(pi/2 + 0.5*phImb*pi/180));
rxSig = imbI + imbQ;

Normalize the power of the received signal

rxSig = rxSig/std(rxSig);

Remove the I/Q imbalance using the comm.IQImbalanceCompensator System object. Set the compensator object such that the complex coefficients are made available as an output argument.

hIQComp = comm.IQImbalanceCompensator('CoefficientOutputPort',true);
[compSig,coef] = step(hIQComp,rxSig);

Estimate the imbalance from the last value of the compensator coefficient.

[ampImbEst,phImbEst] = iqcoef2imbal(coef(end));

Compare the estimated imbalance values with the specified ones. Notice that there is good agreement.

[ampImb phImb; ampImbEst phImbEst]
ans = 2×2

2.0000   15.0000
2.0178   14.5740

## Input Arguments

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Coefficient used to compensate for an I/Q imbalance, specified as a complex-valued vector.

Example: 0.4+0.6i

Example: [0.1+0.2i; 0.3+0.5i]

Data Types: single | double

## Output Arguments

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Amplitude imbalance in dB, returned as a real-valued vector with the same dimensions as C.

Phase imbalance in degrees, returned as a real-valued vector with the same dimensions as C.

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### I/Q Imbalance Compensation

The function iqcoef2imbal is a supporting function for the comm.IQImbalanceCompensator System object™.

Given a scaling and rotation factor, G, compensator coefficient, C, and received signal, x, the compensated signal, y, has the form

$y=G\left[x+C\mathrm{conj}\left(x\right)\right]\text{\hspace{0.17em}}.$

In matrix form, this can be rewritten as

$Y=R\text{ }X\text{\hspace{0.17em}},$

where X is a 2-by-1 vector representing the imbalanced signal [XI, XQ] and Y is a 2-by-1 vector representing the compensator output [YI, YQ].

The matrix R is expressed as

$R=\left[\begin{array}{cc}1+\mathrm{Re}\left\{C\right\}& \mathrm{Im}\left\{C\right\}\\ \mathrm{Im}\left\{C\right\}& 1-\mathrm{Re}\left\{C\right\}\end{array}\right]$

For the compensator to perfectly remove the I/Q imbalance, R = K-1 because $X=K\text{\hspace{0.17em}}S$, where K is a 2-by-2 matrix whose values are determined by the amplitude and phase imbalance and S is the ideal signal. Define a matrix M with the form

$M=\left[\begin{array}{cc}1& -\alpha \\ \alpha & 1\end{array}\right]$

Both M and M-1 can be thought of as scaling and rotation matrices that correspond to the factor G. Because K = R-1, the product M-1 R K M is the identity matrix, where M-1 R represents the compensator output and K M represents the I/Q imbalance. The coefficient α is chosen such that

$K\text{ }M=L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{Q}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{Q}\right)\end{array}\right]$

where L is a constant. From this form, we can obtain Igain, Qgain, θI, and θQ. For a given phase imbalance, ΦImb, the in-phase and quadrature angles can be expressed as

$\begin{array}{c}{\theta }_{I}=-\left(\pi /2\right)\left({\Phi }_{Imb}/180\right)\\ {\theta }_{Q}=\pi /2+\left(\pi /2\right)\left({\Phi }_{Imb}/180\right)\end{array}$

Hence, cos(θQ) = sin(θI) and sin(θQ) = cos(θI) so that

$L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{Q}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{Q}\right)\end{array}\right]=L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{I}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{I}\right)\end{array}\right]$

The I/Q imbalance can be expressed as

$\begin{array}{c}K\text{ }M=\left[\begin{array}{cc}{K}_{11}+\alpha {K}_{12}& -\alpha {K}_{11}+{K}_{12}\\ {K}_{21}+\alpha {K}_{22}& -\alpha {K}_{21}+{K}_{22}\end{array}\right]\\ =L\left[\begin{array}{cc}{I}_{gain}\mathrm{cos}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{sin}\left({\theta }_{I}\right)\\ {I}_{gain}\mathrm{sin}\left({\theta }_{I}\right)& {Q}_{gain}\mathrm{cos}\left({\theta }_{I}\right)\end{array}\right]\end{array}$

Therefore,

$\left({K}_{21}+\alpha {K}_{22}\right)/\left({K}_{11}+\alpha {K}_{12}\right)=\left(-\alpha {K}_{11}+{K}_{12}\right)/\left(-\alpha {K}_{21}+{K}_{22}\right)=\mathrm{sin}\left({\theta }_{I}\right)/\mathrm{cos}\left({\theta }_{I}\right)$

The equation can be written as a quadratic equation to solve for the variable α, that is D1α2 + D2α + D3 = 0, where

$\begin{array}{c}{D}_{1}=-{K}_{11}{K}_{12}+{K}_{22}{K}_{21}\\ {D}_{2}={K}_{12}^{2}+{K}_{21}^{2}-{K}_{11}^{2}-{K}_{22}^{2}\\ {D}_{3}={K}_{11}{K}_{12}-{K}_{21}{K}_{22}\end{array}$

When |C| ≤ 1, the quadratic equation has the following solution:

$\alpha =\frac{-{D}_{2}-\sqrt{{D}^{2}-4{D}_{1}{D}_{3}}}{2{D}_{1}}$

Otherwise, when |C| > 1, the solution has the following form:

$\alpha =\frac{-{D}_{2}+\sqrt{{D}^{2}-4{D}_{1}{D}_{3}}}{2{D}_{1}}$

Finally, the amplitude imbalance, AImb, and the phase imbalance, ΦImb, are obtained.

$\begin{array}{c}{K}^{\prime }=K\left[\begin{array}{cc}1& -\alpha \\ \alpha & 1\end{array}\right]\\ {A}_{Imb}=20{\mathrm{log}}_{10}\left({{K}^{\prime }}_{11}/{{K}^{\prime }}_{22}\right)\\ {\Phi }_{Imb}=-2{\mathrm{tan}}^{-1}\left({{K}^{\prime }}_{21}/{{K}^{\prime }}_{11}\right)\left(180/\pi \right)\end{array}$

### Note

• If C is real and |C| ≤ 1, the phase imbalance is 0 and the amplitude imbalance is 20log10((1–C)/(1+C))

• If C is real and |C| > 1, the phase imbalance is 180° and the amplitude imbalance is 20log10((C+1)/(C−1)).

• If C is imaginary, AImb = 0.