電子教案-高頻電子線路-習(xí)題解答-第二章?一、習(xí)題2.1題目已知載波電壓\(u_c=U_c\cos(\omega_ct)\)，調(diào)制信號(hào)電壓\(u_{\Omega}=U_{\Omega}\cos(\omega_{\Omega}t)\)，試分別寫出調(diào)幅波、調(diào)頻波和調(diào)相波的表達(dá)式。解答1.調(diào)幅波表達(dá)式：普通調(diào)幅波（AM）的表達(dá)式為\(u_{AM}(t)=U_c(1+m_a\cos(\omega_{\Omega}t))\cos(\omega_ct)\)，其中\(zhòng)(m_a=\frac{U_{\Omega}}{U_c}\)為調(diào)幅系數(shù)。展開可得\(u_{AM}(t)=U_c\cos(\omega_ct)+m_aU_c\cos(\omega_ct)\cos(\omega_{\Omega}t)\)。根據(jù)三角函數(shù)的積化和差公式\(\cosA\cosB=\frac{1}{2}[\cos(A+B)+\cos(AB)]\)，進(jìn)一步得到\(u_{AM}(t)=U_c\cos(\omega_ct)+\frac{m_aU_c}{2}[\cos((\omega_c+\omega_{\Omega})t)+\cos((\omega_c\omega_{\Omega})t)]\)。2.調(diào)頻波表達(dá)式：調(diào)頻波的瞬時(shí)頻率\(\omega(t)=\omega_c+k_fu_{\Omega}(t)=\omega_c+k_fU_{\Omega}\cos(\omega_{\Omega}t)\)。設(shè)調(diào)頻波的初始相位為\(\varphi_0\)，則相位\(\varphi(t)=\int_{0}^{t}\omega(\tau)d\tau+\varphi_0=\omega_ct+\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t)+\varphi_0\)。調(diào)頻波的表達(dá)式為\(u_{FM}(t)=U_c\cos(\omega_ct+\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t)+\varphi_0)\)。令\(m_f=\frac{k_fU_{\Omega}}{\omega_{\Omega}}\)為調(diào)頻指數(shù)，則\(u_{FM}(t)=U_c\cos(\omega_ct+m_f\sin(\omega_{\Omega}t)+\varphi_0)\)。3.調(diào)相波表達(dá)式：調(diào)相波的瞬時(shí)相位\(\varphi(t)=\varphi_c+k_pu_{\Omega}(t)=\varphi_c+k_pU_{\Omega}\cos(\omega_{\Omega}t)\)。調(diào)相波的表達(dá)式為\(u_{PM}(t)=U_c\cos(\varphi_c+k_pU_{\Omega}\cos(\omega_{\Omega}t))\)。令\(m_p=k_pU_{\Omega}\)為調(diào)相指數(shù)，則\(u_{PM}(t)=U_c\cos(\varphi_c+m_p\cos(\omega_{\Omega}t))\)。二、習(xí)題2.2題目已知某調(diào)幅波的表達(dá)式為\(u(t)=5(1+0.6\cos(2\pi\times10^3t))\cos(2\pi\times10^6t)\)V，試求：1.載波頻率\(f_c\)、調(diào)制信號(hào)頻率\(f_{\Omega}\)、調(diào)幅系數(shù)\(m_a\)。2.該調(diào)幅波的頻帶寬度\(B_{AM}\)。3.畫出該調(diào)幅波的頻譜圖。解答1.求載波頻率\(f_c\)、調(diào)制信號(hào)頻率\(f_{\Omega}\)、調(diào)幅系數(shù)\(m_a\)：已知調(diào)幅波表達(dá)式\(u(t)=5(1+0.6\cos(2\pi\times10^3t))\cos(2\pi\times10^6t)\)。對(duì)于載波電壓\(u_c=U_c\cos(\omega_ct)\)，這里\(\omega_c=2\pi\times10^6\rad/s\)，所以載波頻率\(f_c=\frac{\omega_c}{2\pi}=10^6\Hz=1\MHz\)。對(duì)于調(diào)制信號(hào)電壓\(u_{\Omega}=U_{\Omega}\cos(\omega_{\Omega}t)\)，這里\(\omega_{\Omega}=2\pi\times10^3\rad/s\)，所以調(diào)制信號(hào)頻率\(f_{\Omega}=\frac{\omega_{\Omega}}{2\pi}=10^3\Hz=1\kHz\)。調(diào)幅系數(shù)\(m_a\)由表達(dá)式中\(zhòng)(\cos(\omega_{\Omega}t)\)的系數(shù)決定，\(m_a=0.6\)。2.求該調(diào)幅波的頻帶寬度\(B_{AM}\)：根據(jù)調(diào)幅波頻帶寬度公式\(B_{AM}=2f_{\Omega}\)。已知\(f_{\Omega}=1\kHz\)，所以\(B_{AM}=2\times1\kHz=2\kHz\)。3.畫出該調(diào)幅波的頻譜圖：調(diào)幅波的頻譜由載波分量和上下邊頻分量組成。載波頻率\(f_c=1\MHz\)，幅度為\(U_c=5\V\)。上邊頻頻率\(f_{c}+f_{\Omega}=1\MHz+1\kHz=1.001\MHz\)，幅度為\(\frac{m_aU_c}{2}=\frac{0.6\times5}{2}=1.5\V\)。下邊頻頻率\(f_{c}f_{\Omega}=1\MHz1\kHz=0.999\MHz\)，幅度為\(\frac{m_aU_c}{2}=1.5\V\)。頻譜圖以頻率\(f\)為橫軸，幅度為縱軸，畫出\(f=0.999\MHz\)、\(f=1\MHz\)、\(f=1.001\MHz\)處的譜線，\(f=1\MHz\)處幅度為\(5\V\)，\(f=0.999\MHz\)和\(f=1.001\MHz\)處幅度為\(1.5\V\)。三、習(xí)題2.3題目已知某調(diào)幅發(fā)射機(jī)，載波功率\(P_c=10\kW\)，調(diào)幅系數(shù)\(m_a=0.8\)，試求：1.邊頻功率\(P_{SB}\)。2.總功率\(P_{T}\)。解答1.求邊頻功率\(P_{SB}\)：調(diào)幅波總功率\(P_T\)與載波功率\(P_c\)和邊頻功率\(P_{SB}\)的關(guān)系為\(P_T=P_c+P_{SB}\)。邊頻功率\(P_{SB}\)的計(jì)算公式為\(P_{SB}=P_c\frac{m_a^2}{2}\)。已知\(P_c=10\kW\)，\(m_a=0.8\)，則\(P_{SB}=10\times\frac{0.8^2}{2}=3.2\kW\)。2.求總功率\(P_{T}\)：由\(P_T=P_c+P_{SB}\)。已求得\(P_c=10\kW\)，\(P_{SB}=3.2\kW\)，所以\(P_T=10+3.2=13.2\kW\)。四、習(xí)題2.4題目已知調(diào)制信號(hào)\(u_{\Omega}(t)=U_{\Omega}\cos(\omega_{\Omega}t)\)，載波電壓\(u_c(t)=U_c\cos(\omega_ct)\)，設(shè)初始相位為零，試推導(dǎo)窄帶調(diào)頻波和窄帶調(diào)相波的表達(dá)式，并說(shuō)明它們之間的關(guān)系。解答1.窄帶調(diào)頻波表達(dá)式推導(dǎo)：調(diào)頻波的瞬時(shí)頻率\(\omega(t)=\omega_c+k_fu_{\Omega}(t)=\omega_c+k_fU_{\Omega}\cos(\omega_{\Omega}t)\)。當(dāng)為窄帶調(diào)頻時(shí)，\(k_fU_{\Omega}\ll\omega_c\)，此時(shí)相位\(\varphi(t)=\int_{0}^{t}\omega(\tau)d\tau=\omega_ct+\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t)\)。窄帶調(diào)頻波表達(dá)式\(u_{NBFM}(t)=U_c\cos(\omega_ct+\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t))\)。根據(jù)三角函數(shù)公式\(\cos(A+B)=\cosA\cosB\sinA\sinB\)，將上式展開得\(u_{NBFM}(t)=U_c\cos(\omega_ct)\cos(\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t))U_c\sin(\omega_ct)\sin(\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t))\)。因?yàn)閈(k_fU_{\Omega}\ll\omega_c\)，\(\cos(\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t))\approx1\)，\(\sin(\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t))\approx\frac{k_fU_{\Omega}}{\omega_{\Omega}}\sin(\omega_{\Omega}t)\)。所以窄帶調(diào)頻波表達(dá)式為\(u_{NBFM}(t)=U_c\cos(\omega_ct)\frac{k_fU_{\Omega}}{\omega_{\Omega}}U_c\sin(\omega_ct)\sin(\omega_{\Omega}t)\)。2.窄帶調(diào)相波表達(dá)式推導(dǎo)：調(diào)相波的瞬時(shí)相位\(\varphi(t)=\varphi_c+k_pu_{\Omega}(t)=\varphi_c+k_pU_{\Omega}\cos(\omega_{\Omega}t)\)。當(dāng)為窄帶調(diào)相時(shí)，\(k_pU_{\Omega}\ll\omega_c\)，窄帶調(diào)相波表達(dá)式\(u_{NBPM}(t)=U_c\cos(\varphi_c+k_pU_{\Omega}\cos(\omega_{\Omega}t))\)。展開得\(u_{NBPM}(t)=U_c\cos(\varphi_c)\cos(k_pU_{\Omega}\cos(\omega_{\Omega}t))U_c\sin(\varphi_c)\sin(k_pU_{\Omega}\cos(\omega_{\Omega}t))\)。因?yàn)閈(k_pU_{\Omega}\ll\omega_c\)，\(\cos(k_pU_{\Omega}\cos(\omega_{\Omega}t))\approx1\)，\(\sin(k_pU_{\Omega}\cos(\omega_{\Omega}t))\approxk_pU_{\Omega}\cos(\omega_{\Omega}t)\)。所以窄帶調(diào)相波表達(dá)式為\(u_{NBPM}(t)=U_c\cos(\varphi_c)k_pU_{\Omega}U_c\sin(\varphi_c)\cos(\omega_{\Omega}t)\)。3.它們之間的關(guān)系：比較窄帶調(diào)頻波和窄帶調(diào)相波表達(dá)式，若令\(k_f=\frac{k_p}{\omega_{\Omega}}\)，則窄帶調(diào)頻波和窄帶調(diào)相波表達(dá)式形式上是相似的。窄帶調(diào)頻波和窄帶調(diào)相波在窄帶情況下，其波形特點(diǎn)和頻譜結(jié)構(gòu)有相似之處，只是調(diào)制指數(shù)與調(diào)制參數(shù)的關(guān)系有所不同。窄帶調(diào)頻的調(diào)制指數(shù)為\(m_f=\frac{k_fU_{\Omega}}{\omega_{\Omega}}\)，窄帶調(diào)相的調(diào)制指數(shù)為\(m_p=k_pU_{\Omega}\)，當(dāng)\(k_f=\frac{k_p}{\omega_{\Omega}}\)時(shí)，兩者調(diào)制指數(shù)在數(shù)值上有一定關(guān)聯(lián)。五、習(xí)題2.5題目已知某調(diào)頻波，載波頻率\(f_c=100\MHz\)，調(diào)制信號(hào)頻率\(f_{\Omega}=1\kHz\)，調(diào)頻指數(shù)\(m_f=10\)，試求：1.調(diào)頻波的頻帶寬度\(B_{FM}\)。2.若調(diào)制信號(hào)幅度不變，頻率增大為\(f_{\Omega}'=2\kHz\)，此時(shí)調(diào)頻波的頻帶寬度\(B_{FM}'\)。解答1.求調(diào)頻波的頻帶寬度\(B_{FM}\)：根據(jù)調(diào)頻波頻帶寬度公式\(B_{FM}=2(m_f+1)f_{\Omega}\)。已知\(f_c=100\MHz\)，\(f_{\Omega}=1\kHz\)，\(m_f=10\)，則\(B_{FM}=2(10+1)\times1\kHz=22\kHz\)。2.求調(diào)制信號(hào)頻率增大后的頻帶寬度\(B_{FM}'\)：當(dāng)調(diào)制信號(hào)頻率變?yōu)閈(f_{\Omega}'=2\kHz\)，調(diào)制信號(hào)幅度不變，調(diào)頻指數(shù)變?yōu)閈(m_f'=\frac{k_fU_{\Omega}}{f_{\Omega}'}\)。由于調(diào)制信號(hào)幅度不變，\(m_f=\frac{k_fU_{\Omega}}{f_{\Omega}}\)，所以\(m_f'=\frac{f_{\Omega}}{f_{\Omega}'}m_f\)。這里\(f_{\Omega}=1\kHz\)，\(f_{\Omega}'=2\kHz\)，\(m_f=10\)，則\(m_f'=\frac{1}{2}\times10=5\)。此時(shí)頻帶寬度\(B_{FM}'=2(m_f'+1)f_{\Omega}'\)。代入\(m_f'=5\)，\(f_{\Omega}'=2\kHz\)，得\(B_{FM}'=2(5+1)\times2\kHz=24\kHz\)。六、習(xí)題2.6題目已知某調(diào)相波，載波頻率\(f_c=50\MHz\)，調(diào)制信號(hào)\(u_{\Omega}(t)=U_{\Omega}\cos(\omega_{\Omega}t)\)，調(diào)相指數(shù)\(m_p=5\)，試求：1.調(diào)制信號(hào)頻率\(f_{\Omega}\)為某值時(shí)調(diào)相波的頻帶寬度\(B_{PM}\)。2.若調(diào)制信號(hào)頻率增大一倍，調(diào)相波的頻帶寬度\(B_{PM}'\)。解答1.求調(diào)制信號(hào)頻率\(f_{\Omega}\)為某值時(shí)調(diào)相波的頻帶寬度