描述某系统的微分方程为y”(t)+4y’(t)+3y(t)=f(t) 求当f(t)=2e-2t，t22650；y(0+)=2，y’(0+)= -1时的全响应

设\(z = {e^{x - 2y}}\)，而\(x = \sin t\)，\(y = {t^3}\)，则全导数\( { { dz} \over {dt}} = \) A: \({e^{\sin t - {t^3}}}(\cos t - 6{t^2})\) B: \({e^{\sin t - 2{t^3}}}(\sin t - 6{t^2})\) C: \({e^{\cos t - 2{t^3}}}(\cos t - 6{t^2})\) D: \({e^{\sin t - 2{t^3}}}(\cos t - 6{t^2})\)

设\(z = {e^{x - 2y}}\)，而\(x = \sin t,\;y = {t^3},\)则\( { { dz} \over {dt}} = \)( ) A: \({e^{\sin t - 2{t^3}}}\) B: \({e^{\sin t - 2{t^3}}}\left( {\cos t - 6{t^2}} \right)\) C: \({e^{\sin t - 2{t^3}}}\ {\sin t } \) D: \({e^{\sin t - 2{t^3}}}\,{t^3}\)

已知有理数t满足|1一t|=1+|t|，则|t一2006|—|1一t|=( )． A: 2 000 B: 2 001 C: 2 002 D: 2 005 E: 2 006

整个睡眠过程经过的时间为 A: 2个t 1/2 B: 3个t 1/2 C: 4个t 1/2 D: 5个t 1/2 E: 6个t 1/2

【多选题】若f 1 (t) = ɛ (-t) , f 2 (t) = e t ,则f 1 (t)* f 2 (t) = A. f 1 ꞌ (t)* f 2 (–1) (t) B. f 1 (–1) (t)* f 2 ꞌ (t) C. f 1 (t-3)* f 2 (t+3) D. f 1 (–3) (t)* f 2 ꞌꞌꞌ (t)

频谱函数F(jw)=2/(jw+3)+4/(jw-2)所对应的信号f(t) A: 2(e)-3t次方u(t)-4(e)2t次方u(t) B: 4(e)2t次方u(t)-2(e)-3t次方u(t) C: 2(e)3t次方u(t)-4(e)-2t次方u(t) D: 4(e)-2t次方u(t)-2(e)3t次方u(t)

以${{e}^{t}}$，$t{{e}^{t}}$为特解的二阶线性常系数齐次微分方程是 A: $\frac{{{d}^{2}}x}{d{{t}^{2}}}-x=0$ B: $\frac{{{d}^{2}}x}{d{{t}^{2}}}-2\frac{dx}{dt}+x=0$ C: $\frac{{{d}^{2}}x}{d{{t}^{2}}}-\frac{dx}{dt}+x=0$ D: $\frac{{{d}^{2}}x}{d{{t}^{2}}}-\frac{dx}{dt}=0$

Fill in the blankFor the expressionf(t)=tε(t)+2ε(t−2)−tε(t−2),f(3)=______ .

【单选题】香农采样定理指出,如果采样器的输入信号e(t) 具有有限带宽,并且有ωm 的最大频率分量,则要使采样信号e*(t) 能不失真地恢复成连续信号e(t) 的采样周期 T ,应满足条件() A. T≤2π/2 ωm B. T≥2π/2 ωm C. T≤2 ωm D. T≥2 ωm