| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine the period of the given signal, we need to find the least common multiple (LCM) of the periods of the individual sinusoidal components. The general formula for the period of a sinusoidal signal is \( T = \frac{2\pi}{\omega} \), where \(\omega\) is the angular frequency. Let's determine the periods of the individual sinusoidal components first: 1. For the first component, \(\cos (3 \pi t)\): - The angular frequency is \(\omega_1 = 3\pi\). - The period is \( T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \). 2. For the second component, \(-4 \cos (5 \pi t-0.5 \pi)\): - The angular frequency is \(\omega_2 = 5\pi\). - The period is \( T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \). Next, we need to find the LCM of \( T_1 = \frac{2}{3} \) and \( T_2 = \frac{2}{5} \). To find the LCM, we first convert the fractions to have a common denominator: - \( T_1 = \frac{2}{3} = \frac{10}{15} \) - \( T_2 = \frac{2}{5} = \frac{6}{15} \) The LCM of the numerators (10 and 6) is 30. Therefore, the LCM of \( T_1 \) and \( T_2 \) is \( \frac{30}{15} = 2 \). Thus, the period of the signal \( x_1(t) \) is \( T = 2 \) seconds. |
2 |
| deepseekr1 | To determine the period of the signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \), we start by finding the periods of each individual cosine term. 1. For the term \( \cos(3\pi t) \): - The angular frequency is \( 3\pi \). - The period \( T_1 \) is calculated as \( \frac{2\pi}{3\pi} = \frac{2}{3} \) seconds. 2. For the term \( -4\cos(5\pi t - 0.5\pi) \): - The angular frequency is \( 5\pi \). - The period \( T_2 \) is calculated as \( \frac{2\pi}{5\pi} = \frac{2}{5} \) seconds. Next, we find the least common multiple (LCM) of the periods \( \frac{2}{3} \) and \( \frac{2}{5} \). To do this, we convert the problem into finding the LCM of the numerators and the greatest common divisor (GCD) of the denominators: - The numerators are 2 and 2, so their LCM is 2. - The denominators are 3 and 5, so their GCD is 1. Thus, the LCM of \( \frac{2}{3} \) and \( \frac{2}{5} \) is \( \frac{2}{1} = 2 \). To verify, we check if \( T = 2 \) satisfies the periodicity condition for both terms: - For \( \cos(3\pi t) \), \( \cos(3\pi(t + 2)) = \cos(3\pi t + 6\pi) = \cos(3\pi t) \). - For \( -4\cos(5\pi t - 0.5\pi) \), \( -4\cos(5\pi(t + 2) - 0.5\pi) = -4\cos(5\pi t + 10\pi - 0.5\pi) = -4\cos(5\pi t - 0.5\pi) \). Both terms confirm that \( T = 2 \) is indeed a period. Checking smaller values (e.g., \( T = 1 \)) shows they do not satisfy the periodicity condition. Thus, the period of the signal is \(\boxed{2}\). |
2 |
| gpto1 | To determine the period of the signal: \[ x_1(t) = \cos(3\pi t) - 4 \cos\left(5\pi t - \frac{\pi}{2}\right) \] we need to find the smallest positive value \( T \) such that \( x_1(t) = x_1(t + T) \) for all \( t \). This overall period depends on the periods of the individual cosine terms. ### Step 1: Find the Periods of the Individual Cosine Functions **First Cosine Term:** The general form of a cosine function is: \[ \cos(\omega t + \phi) \] where \( \omega \) is the angular frequency, and the period \( T \) is given by: \[ T = \frac{2\pi}{\omega} \] For \( \cos(3\pi t) \): - Angular frequency: \( \omega_1 = 3\pi \) - Period: \[ T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \] **Second Cosine Term:** For \( \cos\left(5\pi t - \frac{\pi}{2}\right) \): - Angular frequency: \( \omega_2 = 5\pi \) (the phase shift does not affect the period) - Period: \[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \] ### Step 2: Find the Least Common Multiple (LCM) of the Periods We need to find the smallest \( T \) such that \( T \) is a common multiple of both \( T_1 \) and \( T_2 \): \[ T = n_1 T_1 = n_2 T_2 \] for some positive integers \( n_1 \) and \( n_2 \). Set up the equation: \[ n_1 T_1 = n_2 T_2 \] Substitute \( T_1 = \frac{2}{3} \) and \( T_2 = \frac{2}{5} \): \[ n_1 \left(\frac{2}{3}\right) = n_2 \left(\frac{2}{5}\right) \] Simplify: \[ \frac{2n_1}{3} = \frac{2n_2}{5} \] Multiply both sides by \( \frac{1}{2} \): \[ \frac{n_1}{3} = \frac{n_2}{5} \] Cross-multiply: \[ 5n_1 = 3n_2 \] ### Step 3: Solve for the Smallest Positive Integers We are looking for the smallest positive integers \( n_1 \) and \( n_2 \) that satisfy the equation: \[ 5n_1 = 3n_2 \] This is a Diophantine equation. Let's find the smallest positive integer solutions: - Let \( n_1 = 3 \), then: \[ 5 \times 3 = 15 = 3n_2 \implies n_2 = 5 \] So, \( n_1 = 3 \) and \( n_2 = 5 \). ### Step 4: Calculate the Fundamental Period \( T \) Now, compute \( T \) using either \( n_1 T_1 \) or \( n_2 T_2 \): - Using \( n_1 T_1 \): \[ T = n_1 T_1 = 3 \times \frac{2}{3} = 2 \] - Using \( n_2 T_2 \): \[ T = n_2 T_2 = 5 \times \frac{2}{5} = 2 \] Both calculations yield \( T = 2 \). ### Conclusion The fundamental period of the signal \( x_1(t) \) is: \[ T = 2 \] **Answer:** The period of the signal is **2 units of time**. |
2 |
| GPT-4o | To determine the period of the signal \( x_1(t) = \cos (3 \pi t) - 4 \cos (5 \pi t - 0.5 \pi) \), we need to find the fundamental period of each cosine component and then determine the least common multiple (LCM) of these periods. ### Step-by-Step Solution: 1. **Identify the periods of the individual cosine components:** - For the first term \( \cos(3 \pi t) \): The general form of a cosine function is \( \cos(\omega t + \phi) \), where \( \omega \) is the angular frequency. Here, \( \omega_1 = 3 \pi \). The period \( T_1 \) is given by: \[ T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \] - For the second term \( -4 \cos(5 \pi t - 0.5 \pi) \): Here, \( \omega_2 = 5 \pi \). The period \( T_2 \) is given by: \[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \] 2. **Determine the least common multiple (LCM) of the two periods:** The fundamental period of the combined signal is the LCM of \( T_1 \) and \( T_2 \). - \( T_1 = \frac{2}{3} \) - \( T_2 = \frac{2}{5} \) To find the LCM, we first express \( T_1 \) and \( T_2 \) with a common denominator: \[ T_1 = \frac{2}{3} = \frac{2 \cdot 5}{3 \cdot 5} = \frac{10}{15} \] \[ T_2 = \frac{2}{5} = \frac{2 \cdot 3}{5 \cdot 3} = \frac{6}{15} \] The LCM of \( \frac{10}{15} \) and \( \frac{6}{15} \) is the smallest fraction that is a multiple of both. In this case, we need to find the least common multiple of the numerators 10 and 6, which is 30. Therefore: \[ \text{LCM} \left( \frac{10}{15}, \frac{6}{15} \right) = \frac{30}{15} = 2 \] ### Conclusion: The period of the signal \( x_1(t) = \cos (3 \pi t) - 4 \cos (5 \pi t - 0.5 \pi) \) is \( \boxed{2} \). |
2 |
| llama3.1-70B-Instruct | To determine the period of the given signal, $x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi)$, we need to find the least common multiple (LCM) of the periods of the two cosine functions. The period of $\cos(3\pi t)$ is $\frac{2\pi}{3}$, and the period of $\cos(5\pi t - 0.5\pi)$ is $\frac{2\pi}{5}$. To find the LCM, we can list the multiples of each period: Multiples of $\frac{2\pi}{3}$: $\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{6\pi}{3}, ...$ Multiples of $\frac{2\pi}{5}$: $\frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, ...$ The first common multiple is $\frac{6\pi}{5} \times \frac{5}{3} = 2\pi$. Therefore, the period of the given signal $x_1(t)$ is $2\pi$. |
2 |
| llama3.1-405B-Instruct | The period of a signal is the time interval between two consecutive repetitions of the signal. To determine the period of the given signal, we need to find the time interval between two consecutive repetitions of the signal. The given signal is a sum of two cosine functions: $$x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi)$$ To find the period of the signal, we need to find the period of each cosine function and then find the least common multiple (LCM) of the two periods. The period of the first cosine function is: $$T_1 = \frac{2\pi}{3\pi} = \frac{2}{3}$$ The period of the second cosine function is: $$T_2 = \frac{2\pi}{5\pi} = \frac{2}{5}$$ The LCM of $T_1$ and $T_2$ is: $$T = \text{LCM}\left(\frac{2}{3}, \frac{2}{5}\right) = \frac{2}{3} \times \frac{5}{2} = \frac{5}{3}$$ Therefore, the period of the given signal is $\frac{5}{3}$. |
2 |
| DeepSeek-Coder-V2 | To determine the period of the signal \( x_1(t) = \cos(3 \pi t) - 4 \cos(5 \pi t - 0.5 \pi) \), we need to find the periods of each component of the signal and then determine the least common multiple (LCM) of these periods. 1. **Period of \(\cos(3 \pi t)\)**: - The general form of a cosine function is \(\cos(\omega t)\), where \(\omega\) is the angular frequency. - For \(\cos(3 \pi t)\), the angular frequency \(\omega = 3 \pi\). - The period \(T\) of a cosine function is given by \(T = \frac{2 \pi}{\omega}\). - Therefore, the period of \(\cos(3 \pi t)\) is: \[ T_1 = \frac{2 \pi}{3 \pi} = \frac{2}{3} \] 2. **Period of \(-4 \cos(5 \pi t - 0.5 \pi)\)**: - The general form of a cosine function is \(\cos(\omega t + \phi)\), where \(\omega\) is the angular frequency and \(\phi\) is the phase shift. - For \(-4 \cos(5 \pi t - 0.5 \pi)\), the angular frequency \(\omega = 5 \pi\) and the phase shift \(\phi = -0.5 \pi\). - The period \(T\) of a cosine function is given by \(T = \frac{2 \pi}{\omega}\). - Therefore, the period of \(-4 \cos(5 \pi t - 0.5 \pi)\) is: \[ T_2 = \frac{2 \pi}{5 \pi} = \frac{2}{5} \] 3. **Finding the least common multiple (LCM) of \(T_1\) and \(T_2\)**: - We need to find the LCM of \(\frac{2}{3}\) and \(\frac{2}{5}\). - First, find the LCM of the numerators 2 and 2, which is 2. - Then, find the LCM of the denominators 3 and 5, which is 15. - Therefore, the LCM of \(\frac{2}{3}\) and \(\frac{2}{5}\) is: \[ \text{LCM} = \frac{\text{LCM of numerators}}{\text{LCM of denominators}} = \frac{2}{15} \] Thus, the period of the signal \( x_1(t) = \cos(3 \pi t) - 4 \cos(5 \pi t - 0.5 \pi) \) is \(\frac{2}{15}\). |
2 |
| Qwen2-72B-Instruct | To determine the period of the given signal, we need to find the least common multiple (LCM) of the periods of the individual sinusoidal components. The general formula for the period of a sinusoidal signal is $T = \frac{2\pi}{\omega}$, where $\omega$ is the angular frequency. Let's determine the periods of the individual sinusoidal components first: 1. For the first component, $\cos (3 \pi t)$: - The angular frequency is $\omega_1 = 3\pi$. - The period is $T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3}$. 2. For the second component, $-4 \cos (5 \pi t-0.5 \pi)$: - The angular frequency is $\omega_2 = 5\pi$. - The period is $T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5}$. Next, we need to find the LCM of $T_1 = \frac{2}{3}$ and $T_2 = \frac{2}{5}$. To find the LCM, we first convert the fractions to have a common denominator: - $T_1 = \frac{2}{3} = \frac{10}{15}$ - $T_2 = \frac{2}{5} = \frac{6}{15}$ The LCM of the numerators (10 and 6) is 30. Therefore, the LCM of $T_1$ and $T_2$ is $\frac{30}{15} = 2$. Thus, the period of the signal $x_1(t)$ is $T = 2$ seconds. |
2 |
| baichuan4 | To determine the period of the signal $$x_1(t)=\cos (3 \pi t)-4 \cos (5 \pi t-0.5 \pi)$$, we need to find the smallest positive value of T such that $$x_1(t)=x_1(t+T)$$ for all t. The period of a cosine function is given by $$\frac{2\pi}{\omega}$$, where $$\omega$$ is the angular frequency. For the first term, $$\cos (3 \pi t)$$, the angular frequency is $$3\pi$$, so its period is $$\frac{2\pi}{3\pi}=\frac{2}{3}$$. For the second term, $$-4\cos (5 \pi t-0.5 \pi)$$, the angular frequency is $$5\pi$$, so its period is $$\frac{2\pi}{5\pi}=\frac{2}{5}$$. Since the periods of both terms are different, the period of the combined signal $$x_1(t)$$ is the least common multiple (LCM) of $$\frac{2}{3}$$ and $$\frac{2}{5}$$. The LCM of $$\frac{2}{3}$$ and $$\frac{2}{5}$$ is $$\frac{2}{1}$$ (or simply 2), because 3 and 5 are coprime. Therefore, the period of $$x_1(t)$$ is 2. |
2 |
| doubao | The period of a sinusoidal signal of the form $\cos(\omega t)$ is $T = \frac{2\pi}{\omega}$. For the first term, $\cos(3\pi t)$, the angular frequency is $3\pi$, so the period is $T_1 = \frac{2\pi}{3\pi} = \frac{2}{3}$. For the second term, $-4\cos(5\pi t - 0.5\pi)$, the angular frequency is $5\pi$, so the period is $T_2 = \frac{2\pi}{5\pi} = \frac{2}{5}$. The overall period of the signal is the least common multiple (LCM) of $T_1$ and $T_2$. The LCM of $\frac{2}{3}$ and $\frac{2}{5}$ is $\frac{10}{1} = 10$. So the period of the signal $x_1(t)$ is 10. |
2 |
| minimax | To determine the period of the given signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \), we need to find the period of each individual cosine function and then find the least common multiple (LCM) of these periods, as the period of the sum of two or more periodic functions is the LCM of their individual periods. The general form of a cosine function is \( \cos(2\pi f t + \phi) \), where \( f \) is the frequency in hertz (Hz), \( t \) is the time in seconds, and \( \phi \) is the phase shift in radians. For the first term, \( \cos(3\pi t) \), the frequency \( f_1 \) is \( \frac{3\pi}{2\pi} = 1.5 \) Hz. The period \( T_1 \) of this function is the reciprocal of the frequency: \[ T_1 = \frac{1}{f_1} = \frac{1}{1.5} = \frac{2}{3} \text{ seconds} \] For the second term, \( -4\cos(5\pi t - 0.5\pi) \), the frequency \( f_2 \) is \( \frac{5\pi}{2\pi} = 2.5 \) Hz. The period \( T_2 \) of this function is: \[ T_2 = \frac{1}{f_2} = \frac{1}{2.5} = \frac{2}{5} \text{ seconds} \] Now, we need to find the LCM of \( T_1 \) and \( T_2 \). To do this, we can express each period as a fraction of the smallest common unit, which in this case is seconds. The LCM of \( \frac{2}{3} \) and \( \frac{2}{5} \) can be found by first finding the LCM of the denominators (3 and 5) and then ensuring that the resulting fraction can be expressed in terms of the original periods. Since 3 and 5 are both prime numbers, their LCM is simply their product, which is 15. Now we need to express \( \frac{2}{3} \) and \( \frac{2}{5} \) in terms of 15ths: \[ T_1 = \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \] \[ T_2 = \frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \] The LCM of 10 and 6 is 30, so the LCM of \( T_1 \) and \( T_2 \) is \( \frac{30}{15} \) seconds, which simplifies to 2 seconds. Therefore, the period of the signal \( x_1(t) \) is 2 seconds. |
2 |
| kimi | To determine the period of the given signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \), we need to find the period of each cosine function separately and then find the least common multiple (LCM) of those periods. 1. The first term is \( \cos(3\pi t) \). The general form of a cosine function is \( \cos(\omega t) \), where \( \omega \) is the angular frequency. The period \( T \) of a cosine function is given by \( T = \frac{2\pi}{\omega} \). For \( \cos(3\pi t) \), \( \omega = 3\pi \), so the period \( T_1 \) is: \[ T_1 = \frac{2\pi}{3\pi} = \frac{2}{3} \] 2. The second term is \( \cos(5\pi t - 0.5\pi) \). The cosine function is periodic with respect to its argument, so a phase shift does not affect the period. The period of \( \cos(5\pi t) \) is: \[ T_2 = \frac{2\pi}{5\pi} = \frac{2}{5} \] Now, we need to find the LCM of \( \frac{2}{3} \) and \( \frac{2}{5} \) to determine the period of the entire signal. The LCM of two numbers is the smallest number that is a multiple of both numbers. To find the LCM of fractions, we can convert them to have a common denominator and then find the LCM of the numerators. The common denominator for 3 and 5 is 15. So we convert the fractions: \[ \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \] \[ \frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \] Now, we find the LCM of 10 and 6, which is 30. This means the LCM of \( \frac{2}{3} \) and \( \frac{2}{5} \) is \( \frac{30}{15} \), which simplifies to 2. Therefore, the period of the signal \( x_1(t) \) is \( T = 2 \) seconds. |
2 |
| wenxinyiyan | 为了确定信号$x_1(t)=\cos (3 \pi t)-4 \cos (5 \pi t-0.5 \pi)$的周期，我们需要分别考虑两个余弦项的周期，并找到它们的最小公倍数。 第一个余弦项是$\cos (3 \pi t)$。余弦函数的基本周期是$2\pi$，因此，该项的周期$T_1$可以通过将基本周期除以角频率$3\pi$来得到： $$ T_1 = \frac{2\pi}{3\pi} = \frac{2}{3} $$ 第二个余弦项是$-4 \cos (5 \pi t-0.5 \pi)$。同样地，该项的周期$T_2$可以通过将基本周期$2\pi$除以角频率$5\pi$来得到： $$ T_2 = \frac{2\pi}{5\pi} = \frac{2}{5} $$ 现在，我们需要找到$T_1$和$T_2$的最小公倍数（LCM），以确定整个信号$x_1(t)$的周期。 $T_1 = \frac{2}{3}$和$T_2 = \frac{2}{5}$的最小公倍数是它们分母的最小公倍数除以它们分子的最大公约数。 3和5的最小公倍数是15，2的最大公约数是2，因此： $$ LCM\left(\frac{2}{3}, \frac{2}{5}\right) = \frac{LCM(3,5)}{GCD(2,2)} = \frac{15}{2} = 1.5 $$ 所以，信号$x_1(t)$的周期是1.5。 |
2 |
| zhipu | To determine the period of the signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \), we need to find the fundamental period of each component and then determine the least common multiple (LCM) of these periods. ### Step 1: Identify the periods of the individual components 1. **First component: \( \cos(3\pi t) \)** - The general form of a cosine function is \( \cos(\omega t) \), where \( \omega \) is the angular frequency. - For \( \cos(3\pi t) \), the angular frequency \( \omega_1 = 3\pi \). - The period \( T_1 \) of this component is given by: \[ T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \] 2. **Second component: \( -4\cos(5\pi t - 0.5\pi) \)** - The phase shift \( -0.5\pi \) does not affect the period of the cosine function. - For \( \cos(5\pi t) \), the angular frequency \( \omega_2 = 5\pi \). - The period \( T_2 \) of this component is given by: \[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \] ### Step 2: Determine the least common multiple (LCM) of the periods To find the period of the combined signal \( x_1(t) \), we need the least common multiple of \( T_1 \) and \( T_2 \). - \( T_1 = \frac{2}{3} \) - \( T_2 = \frac{2}{5} \) First, express these periods in terms of a common numerator: - \( T_1 = \frac{2}{3} = \frac{2 \times 5}{3 \times 5} = \frac{10}{15} \) - \( T_2 = \frac{2}{5} = \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \) The LCM of the denominators (15) is already aligned, so we need the LCM of the numerators (10 and 6). - The LCM of 10 and 6 is 30. Thus, the LCM of \( \frac{10}{15} \) and \( \frac{6}{15} \) is \( \frac{30}{15} = 2 \). ### Conclusion The period of the signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \) is \( T = 2 \). Therefore, the period of the given signal is \( \boxed{2} \). |
2 |
| gpto1 | To determine the period of the signal: \[ x_1(t) = \cos(3\pi t) - 4 \cos\left(5\pi t - \frac{\pi}{2}\right) \] we need to find the smallest positive value \( T \) such that \( x_1(t) = x_1(t + T) \) for all \( t \). This overall period depends on the periods of the individual cosine terms. ### Step 1: Find the Periods of the Individual Cosine Functions **First Cosine Term:** The general form of a cosine function is: \[ \cos(\omega t + \phi) \] where \( \omega \) is the angular frequency, and the period \( T \) is given by: \[ T = \frac{2\pi}{\omega} \] For \( \cos(3\pi t) \): - Angular frequency: \( \omega_1 = 3\pi \) - Period: \[ T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \] **Second Cosine Term:** For \( \cos\left(5\pi t - \frac{\pi}{2}\right) \): - Angular frequency: \( \omega_2 = 5\pi \) (the phase shift does not affect the period) - Period: \[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \] ### Step 2: Find the Least Common Multiple (LCM) of the Periods We need to find the smallest \( T \) such that \( T \) is a common multiple of both \( T_1 \) and \( T_2 \): \[ T = n_1 T_1 = n_2 T_2 \] for some positive integers \( n_1 \) and \( n_2 \). Set up the equation: \[ n_1 T_1 = n_2 T_2 \] Substitute \( T_1 = \frac{2}{3} \) and \( T_2 = \frac{2}{5} \): \[ n_1 \left(\frac{2}{3}\right) = n_2 \left(\frac{2}{5}\right) \] Simplify: \[ \frac{2n_1}{3} = \frac{2n_2}{5} \] Multiply both sides by \( \frac{1}{2} \): \[ \frac{n_1}{3} = \frac{n_2}{5} \] Cross-multiply: \[ 5n_1 = 3n_2 \] ### Step 3: Solve for the Smallest Positive Integers We are looking for the smallest positive integers \( n_1 \) and \( n_2 \) that satisfy the equation: \[ 5n_1 = 3n_2 \] This is a Diophantine equation. Let's find the smallest positive integer solutions: - Let \( n_1 = 3 \), then: \[ 5 \times 3 = 15 = 3n_2 \implies n_2 = 5 \] So, \( n_1 = 3 \) and \( n_2 = 5 \). ### Step 4: Calculate the Fundamental Period \( T \) Now, compute \( T \) using either \( n_1 T_1 \) or \( n_2 T_2 \): - Using \( n_1 T_1 \): \[ T = n_1 T_1 = 3 \times \frac{2}{3} = 2 \] - Using \( n_2 T_2 \): \[ T = n_2 T_2 = 5 \times \frac{2}{5} = 2 \] Both calculations yield \( T = 2 \). ### Conclusion The fundamental period of the signal \( x_1(t) \) is: \[ T = 2 \] **Answer:** The period of the signal is **2 units of time**. |
2 |
| deepseekv3 | To determine the period of the signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \), we need to find the period of each component and then determine the least common multiple (LCM) of these periods. ### Step 1: Find the period of each cosine component 1. **First component**: \( \cos(3\pi t) \) - The general form of a cosine function is \( \cos(\omega t) \), where \( \omega \) is the angular frequency. - Here, \( \omega_1 = 3\pi \). - The period \( T_1 \) is given by: \[ T_1 = \frac{2\pi}{\omega_1} = \frac{2\pi}{3\pi} = \frac{2}{3} \] 2. **Second component**: \( \cos(5\pi t - 0.5\pi) \) - Here, \( \omega_2 = 5\pi \). - The period \( T_2 \) is given by: \[ T_2 = \frac{2\pi}{\omega_2} = \frac{2\pi}{5\pi} = \frac{2}{5} \] ### Step 2: Determine the least common multiple (LCM) of the periods To find the period \( T \) of the composite signal \( x_1(t) \), we need to find the smallest time \( T \) such that both \( T_1 \) and \( T_2 \) divide \( T \) exactly. - \( T_1 = \frac{2}{3} \) - \( T_2 = \frac{2}{5} \) The LCM of \( T_1 \) and \( T_2 \) is the smallest number that is a multiple of both \( T_1 \) and \( T_2 \). To find this, we can use the formula: \[ \text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{GCD}(b, d)} \] Where: - \( \text{LCM} \) is the least common multiple - \( \text{GCD} \) is the greatest common divisor Applying this to our periods: - \( T_1 = \frac{2}{3} \) - \( T_2 = \frac{2}{5} \) - \( \text{LCM}(2, 2) = 2 \) - \( \text{GCD}(3, 5) = 1 \) So, the LCM of \( T_1 \) and \( T_2 \) is: \[ T = \frac{2}{1} = 2 \] ### Conclusion The period of the signal \( x_1(t) = \cos(3\pi t) - 4\cos(5\pi t - 0.5\pi) \) is \( T = 2 \). |
2 |