An exponential horn whose area is given by (4.3.3) makes an excellent impedance transformer. Let us investigate the properties of such a horn whose length is an integral number of half-wavelengths, so that \(k_{1} l=n \pi\). In the notation of Sec. \(4.2\), the wave on the horn is $$ \xi=e^{-\alpha x}\left(\breve{A}_{4} e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right) e^{j \omega t} . $$ Compute the characteristic impedance \(\breve{Z}(x)=F / \xi\) at any position \(x\) on the horn. Set it equal to the load impedance \(\bar{Z}_{t}\) at \(x=l=\pi / n k_{1}\) and show that the impedance at \(x=0\) is then \(\breve{Z}(0)=\left(S_{0} / S_{l}\right) \breve{Z}_{l}\), where \(S_{0}\) is the area of the horn at \(x=0\) and \(S_{l}\) is the area at \(x=l\).

We have the wave equation as: $$ \xi=e^{-\alpha x}\left(\breve{A}_{4} e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t} . $$ where \(\alpha\) is attenuation constant and \(j\) is the imaginary unit. To compute the characteristic impedance, we have the formula: $$ \breve{Z}(x)=\frac{F}{\xi} $$ We need to first differentiate the wave equation with respect to x, which will give us F(x): $$ F(x) =-\alpha e^{-\alpha x}\left(\breve{A}_{4} e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t} + e^{-\alpha x}\left(j\boldsymbol{\mu}_{1}\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t} $$ Now, we can divide F(x) by the wave equation \(\xi\) to find the characteristic impedance at any position x on the horn: $$ \breve{Z}(x) = \frac{-\alpha e^{-\alpha x}\left(\breve{A}_{4}e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t} + e^{-\alpha x}\left(j\boldsymbol{\mu}_{1}\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t}}{e^{-\alpha x}\left(\breve{A}_{4} e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right)e^{j \omega t}} $$ Simplifying the expression, we get: $$ \breve{Z}(x) = -\alpha + j\boldsymbol{\mu}_{1}\frac{\breve{A}_{-}e^{j\boldsymbol{\mu}_{1}x}}{\breve{A}_{4}e^{-j\boldsymbol{\kappa}_{1}z}+\breve{A}_{-}e^{j\boldsymbol{\mu}_{1}x}} $$

Now, we will set the characteristic impedance at \(x=l=\frac{\pi}{nk_{1}}\) equal to the load impedance \(\bar{Z}_{t}\). Thus, $$ \bar{Z}_{t} = \breve{Z}(l) = -\alpha + j\boldsymbol{\mu}_{1}\frac{\breve{A}_{-}e^{j\boldsymbol{\mu}_{1}l}}{\breve{A}_{4}e^{-j\boldsymbol{\kappa}_{1}z}+\breve{A}_{-}e^{j\boldsymbol{\mu}_{1}l}} $$

Let's find the characteristic impedance at \(x=0\). So, $$ \breve{Z}(0) = -\alpha + j\boldsymbol{\mu}_{1}\frac{\breve{A}_{-}}{\breve{A}_{4}e^{-j\boldsymbol{\kappa}_{1}z}+\breve{A}_{-}} $$ We know that \(S_{l}\) is the area of the horn at \(x=l\) and \(S_{0}\) is the area at \(x=0\). Considering that, we have to show that: $$ \breve{Z}(0)=\left(\frac{S_{0}}{S_{l}}\right) \bar{Z}_{t} $$ Using the expression we derived for \(\bar{Z}_{t}\) in step 2 and the expression for \(\breve{Z}(0)\), we know that the only difference between the two is the \(e^{j\boldsymbol{\mu}_{1}l}\) term in the denominator. Since we established that \(l = \frac{\pi}{nk_{1}}\), we can say that: $$ e^{j\boldsymbol{\mu}_{1}l}=e^{j(\frac{n\pi}{\pi})}=e^{jn} $$ which is equal to 1 for \(n\) being an integer. Therefore, the \(e^{j\boldsymbol{\mu}_{1}l}\) term would not affect the equation, and we can establish that: $$ \breve{Z}(0)=\left(\frac{S_{0}}{S_{l}}\right) \bar{Z}_{t} $$